Thu. Jul 18th, 2024

Koshe dekhi 9.3 class 10

Koshe dekhi 9.3 class 10

Q1. (a)  \fn_cm {\color{Blue} m+\frac{1}{m}=\sqrt{3}}  হলে, 

(i) \fn_cm {\color{Blue} {{m}^{2}}+\frac{1}{{{m}^{2}}}}

(ii) \fn_cm {\color{Blue} {{m}^{3}}+\frac{1}{{{m}^{3}}}}

এদের সরলতম মান নির্ণয় করি।

সমাধান (i) :

(i) \fn_cm {{m}^{2}}+\frac{1}{{{m}^{2}}}

\fn_cm =\left ( m+\frac{1}{m} \right )^{2}-2\times m\times \frac{1}{m}

\fn_cm =\left ( \sqrt{3} \right )^{2}-2\: \: {\color{Blue} \left [ \because m+\frac{1}{m}=\sqrt{3} \right ]}

\fn_cm =3-2

\fn_cm =1

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{m}^{2}}+\frac{1}{{{m}^{2}}}} এর সরলতম মান  1

 

সমাধান (ii) :

(ii) \fn_cm m^{3}+\frac{1}{m^{3}}

\fn_cm =\left ( m+\frac{1}{m} \right )^{3}-3\times m\times \frac{1}{m}\left ( m+\frac{1}{m} \right )

\fn_cm =\left ( \sqrt{3} \right )^{3}-3\times \sqrt{3}\: \: {\color{Blue} \left [ \because m+\frac{1}{m}=\sqrt{3} \right ]}

\fn_cm =3\sqrt{3}-3\sqrt{3}

\fn_cm =0

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{m}^{3}}+\frac{1}{{{m}^{3}}}} এর সরলতম মান  0

 

Q1. (b) দেখাই যে, \fn_cm \small {\color{Blue} \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}}

সমাধানঃ

বামপক্ষ \fn_cm \small =\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}

\fn_cm \small =\frac{\left ( \sqrt{5}+\sqrt{3} \right )^{2}-\left ( \sqrt{5}-\sqrt{3} \right )^{2}}{\left ( \sqrt{5}-\sqrt{3} \right )\left ( \sqrt{5}+\sqrt{3} \right )}

\small =\frac{\left [\left ( \sqrt{5} \right )^{2}+2\times \sqrt{5}\times \sqrt{3}+\left ( \sqrt{3} \right )^{2} \right ]-\left [\left ( \sqrt{5} \right )^{2}-2\times \sqrt{5}\times \sqrt{3}+\left ( \sqrt{3} \right )^{2} \right ]}{\left ( \sqrt{5} \right )^{2}-\left ( \sqrt{3} \right )^{2}}\fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_cm =\frac{5+2\sqrt{15}+3-5+2\sqrt{15}-3}{5-3}

\fn_cm =\frac{4\sqrt{15}}{2}

\fn_cm =2\sqrt{15}= ডানপক্ষ ( প্রমাণিত )

 

Q2. সরল করি :

(a) \fn_cm {\color{Blue} \frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}}

সমাধানঃ 

\fn_cm \frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}

\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [\frac{\left( 2+\sqrt{3} \right)}{\left( \sqrt{3}+1 \right)}-\frac{\left( 2-\sqrt{3} \right)}{\left( \sqrt{3}-1 \right)} \right ]

\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [\frac{\left ( 2+\sqrt{3} \right )\left ( \sqrt{3}-1 \right )-\left ( 2-\sqrt{3} \right )\left ( \sqrt{3}+1 \right )}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3}-1 \right )} \right ]

\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [ \frac{\left \{ 2 \sqrt{3}-2+3-\sqrt{3}\right \}-\left \{ 2 \sqrt{3}+2-3-\sqrt{3} \right \}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}} \right ]\; \; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [ \frac{2\sqrt{3}-2+3-\sqrt{3}-2\sqrt{3}-2+3+\sqrt{3}}{3-1} \right ]

\fn_phv =\frac{\sqrt{2}}{\sqrt{3}}\left [ \frac{2}{2} \right ]

\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}

\fn_cm =\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} [ হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{\sqrt{6}}{3}

উত্তরঃ নির্ণেয় সরলতম মান \fn_cm {\color{DarkGreen} \frac{\sqrt{6}}{3}}

 

Q2. সরল করি :

(b) \fn_cm {\color{Blue} \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}}

সমাধানঃ 

\fn_cm \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}

\fn_cm =\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{\left ( \sqrt{5}+\sqrt{2} \right )\left ( \sqrt{5}-\sqrt{2} \right )}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{\left ( \sqrt{2}+\sqrt{7} \right )\left ( \sqrt{2}-\sqrt{7} \right )}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{\left ( \sqrt{7}+\sqrt{5} \right )\left ( \sqrt{7}-\sqrt{5} \right )} [হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{\left ( \sqrt{5} \right )^{2}-\left ( \sqrt{2} \right )^{2}}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{\left ( \sqrt{2} \right )^{2}-\left ( \sqrt{7} \right )^{2}}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{5} \right )^{2}}

=\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{5-2}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{2-7}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{7-5}

=\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{3}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{-5}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{2}

\fn_cm =\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )+\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )+\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )

\fn_cm =\sqrt{35}-\sqrt{14}+\sqrt{10}-\sqrt{35}+\sqrt{14}-\sqrt{10}

\fn_cm =0

উত্তরঃ নির্ণেয় সরলতম মান \fn_jvn {\color{DarkGreen} 0}

 

Q2. সরল করি :

(c) \fn_cm {\color{Blue} \frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}}

সমাধানঃ 

\fn_cm \frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}

\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{\left ( 2-\sqrt{2} \right )\left ( 2+\sqrt{2} \right )}-\frac{30\left ( 4\sqrt{3}+\sqrt{18} \right )}{\left ( 4\sqrt{3}-\sqrt{18} \right )\left ( 4\sqrt{3}+\sqrt{18} \right )}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{\left ( 3-\sqrt{12} \right )\left ( 3+\sqrt{12} \right )}  [ হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{\left ( 2 \right )^{2}-\left ( \sqrt{2} \right )^{2}}-\frac{30\left ( 4 \sqrt{3}+\sqrt{18} \right )}{\left ( 4\sqrt{3} \right )^{2}-\left ( \sqrt{18} \right )^{2}}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{\left ( 3 \right )^{2}-\left ( \sqrt{12} \right )^{2}}\; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{4-2}-\frac{30\left ( 4 \sqrt{3}+\sqrt{18} \right )}{48-18}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{9-12}

\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{2}-\frac{30\left ( 4\sqrt{3}+3\sqrt{2} \right )}{30}-\frac{3\sqrt{2}\left ( 3+2\sqrt{3} \right )}{-3}\; {\color{Blue} \left [ \because \sqrt{18}=3\sqrt{2} ,\; \sqrt{12}=2\sqrt{3}\right ]}

\fn_cm =2\sqrt{3}\left ( 2+\sqrt{2} \right )-\left ( 4\sqrt{3}+3\sqrt{2} \right )+\sqrt{2}\left ( 3+2\sqrt{3} \right )

\fn_cm =4\sqrt{6}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}

\fn_cm =4\sqrt{6}

উত্তরঃ নির্ণেয় সরলতম মান \fn_cm {\color{DarkGreen} 4\sqrt{6}}

Koshe dekhi 9.3 class 10

Q2. সরল করি :

(d) \fn_cm {\color{Blue} \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}}

সমাধানঃ 

\fn_cm \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left (\sqrt{3}+\sqrt{6} \right )\left (\sqrt{3}-\sqrt{6} \right )}-\frac{4\sqrt{3}\left (\sqrt{6}-\sqrt{2} \right )}{\left (\sqrt{6}+\sqrt{2} \right )\left (\sqrt{6}-\sqrt{2} \right )}+\frac{\sqrt{6}\left (\sqrt{2}-\sqrt{3} \right )}{\left (\sqrt{2}+\sqrt{3} \right )\left (\sqrt{2}-\sqrt{3} \right )} [ হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{6} \right )^{2}}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{\left ( \sqrt{6} \right )^{2}-\left ( \sqrt{2} \right )^{2}}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{\left ( \sqrt{2} \right )^{2}-\left ( \sqrt{3} \right )^{2}}\; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{3-6}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{6-2}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{2-3}

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{-3}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{4}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{-1}

\fn_cm =-\sqrt{2}\left ( \sqrt{3}-\sqrt{6} \right )-\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )-\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )

\fn_cm =-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}

\fn_cm =0

উত্তরঃ নির্ণেয় সরলতম মান 0

Koshe dekhi 9.3 class 10

Q3. যদি  \fn_cm {\color{Blue} x=2,y=3}  এবং   \fn_cm {\color{Blue} z=6}  হয়, তবে,  \small {\color{Blue} \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}  এর মান হিসাব করে লিখি।

সমাধানঃ 

\fn_cm \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}

\fn_cm =\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} [ x , y ,z  এর মান বসিয়ে পাই ]

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left (\sqrt{3}+\sqrt{6} \right )\left (\sqrt{3}-\sqrt{6} \right )}-\frac{4\sqrt{3}\left (\sqrt{6}-\sqrt{2} \right )}{\left (\sqrt{6}+\sqrt{2} \right )\left (\sqrt{6}-\sqrt{2} \right )}+\frac{\sqrt{6}\left (\sqrt{2}-\sqrt{3} \right )}{\left (\sqrt{2}+\sqrt{3} \right )\left (\sqrt{2}-\sqrt{3} \right )} [ হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{6} \right )^{2}}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{\left ( \sqrt{6} \right )^{2}-\left ( \sqrt{2} \right )^{2}}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{\left ( \sqrt{2} \right )^{2}-\left ( \sqrt{3} \right )^{2}}\; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{3-6}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{6-2}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{2-3}

\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{-3}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{4}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{-1}

\fn_cm =-\sqrt{2}\left ( \sqrt{3}-\sqrt{6} \right )-\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )-\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )

\fn_cm =-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}

\fn_cm =0

উত্তরঃ নির্ণেয়   \fn_cm {\color{DarkGreen} \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}  এর মান \fn_jvn {\color{DarkGreen} 0} .

Koshe dekhi 9.3 class 10

Q4.  \fn_cm {\color{Blue} x=\sqrt{7}+\sqrt{6}}  হলে,

(i) \fn_cm {\color{Blue} x-\frac{1}{x}}

(ii) \fn_cm {\color{Blue} x+\frac{1}{x}}

(iii) \fn_cm {\color{Blue} {{x}^{2}}+\frac{1}{{{x}^{2}}}}

(iv) \fn_cm {\color{Blue} {{x}^{3}}+\frac{1}{{{x}^{3}}}}

এদের সরলতম মান নির্ণয় করি।

সমাধানঃ 

\fn_cm \because x=\left ( \sqrt{7} +\sqrt{6}\right )

\fn_cm \therefore \frac{1}{x}=\frac{1}{\left ( \sqrt{7}+\sqrt{6} \right )}

\fn_cm =\frac{1}{\left ( \sqrt{7}+\sqrt{6} \right )}\times \frac{\left ( \sqrt{7}-\sqrt{6} \right )}{\left ( \sqrt{7}-\sqrt{6} \right )} [ হরের করণী নিরসন করে পাই ]

\fn_cm =\frac{\left ( \sqrt{7}-\sqrt{6} \right )}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{6} \right )^{2}}

\fn_cm =\frac{\left ( \sqrt{7}-\sqrt{6} \right )}{7-6}

\fn_cm =\left ( \sqrt{7}-\sqrt{6} \right )

 

সমাধান (i) :

এখন, (i) \fn_cm x-\frac{1}{x}

\fn_jvn =\left ( \sqrt{7}+\sqrt{6} \right )-\left ( \sqrt{7}-\sqrt{6} \right ) [ x\fn_cm {\color{Blue} \frac{1}{x}} এর মান  বসিয়ে পাই ]

\fn_cm =\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6}

\fn_cm =2\sqrt{6}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} x-\frac{1}{x}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 2\sqrt{6}}

 

সমাধান (ii) :

(ii) \fn_cm x+\frac{1}{x}

\fn_cm =\left ( \sqrt{7}+\sqrt{6} \right )+\left ( \sqrt{7}-\sqrt{6} \right ) [ x\fn_cm {\color{Blue} \frac{1}{x}} এর মান  বসিয়ে পাই ]

\fn_cm =\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6}

\fn_cm =2\sqrt{7}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} x+\frac{1}{x}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 2\sqrt{7}}

 

সমাধান (iii) :

(iii) \fn_cm {{x}^{2}}+\frac{1}{{{x}^{2}}}

\fn_cm =\left ( x+\frac{1}{x} \right )^{2}-2\cdot x\cdot \frac{1}{x}

\fn_cm =\left ( 2\sqrt{7} \right )^{2}-2    [ \fn_cm {\color{Blue} x+\frac{1}{x}} এর মান বসিয়ে পাই ]

\fn_cm =28-2

\fn_cm =26

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{x}^{2}}+\frac{1}{{{x}^{2}}}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 26}

 

 

সমাধান (iv) :

(iv) \fn_cm {{x}^{3}}+\frac{1}{{{x}^{3}}}

\fn_cm =\left ( x+\frac{1}{x} \right )^{3}-3\cdot x\cdot \frac{1}{x}\cdot \left ( x+\frac{1}{x} \right )

\fn_cm =\left ( 2\sqrt{7} \right )^{3}-3\times 2\sqrt{7}      [ \fn_cm {\color{Blue} x+\frac{1}{x}} এর মান বসিয়ে পাই ]

\fn_cm =56\sqrt{7}-6\sqrt{7}

\fn_cm =50\sqrt{7}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{x}^{3}}+\frac{1}{{{x}^{3}}}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 50\sqrt{7}}.

Koshe dekhi 9.3 class 10

Q5. সরল করি :  \small {\color{Blue} \frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}}  সরলফল \fn_cm {\color{Blue} 14} হলে, \fn_cm {\color{Blue} x} -এর মান কী কী হবে হিসাব করে লিখি।

সমাধানঃ 

\fn_cm \frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}=14

বা, \fn_cm \frac{\left ( x+\sqrt{x^{2}-1} \right )^{2}+\left ( x-\sqrt{x^{2}-1} \right )^{2}}{\left ( x-\sqrt{x^{2}-1} \right )\left ( x+\sqrt{x^{2}-1} \right )}=14

বা, \fn_cm \frac{2\left [\left ( x\right )^{2} +\left ( \sqrt{x^{2}-1} \right ) ^{2}\right ]}{\left ( x\right )^{2} -\left ( \sqrt{x^{2}-1} \right ) ^{2}}=14  \fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}

বা, \fn_cm \frac{2\left ( x^{2} + x^{2}-1\right )}{ x^{2}- x^{2}+1}=14

বা, \fn_cm \frac{2\left ( 2x^{2}-1 \right )}{1}=14

বা, \fn_cm 2x^{2}-1=\frac{14}{2}

বা, \fn_cm 2x^{2}=7+1

বা, \fn_cm 2x^{2}=8

বা, \fn_cm x^{2}=\frac{8}{2}=4

বা, \fn_cm x=\pm \sqrt{4}

\fn_cm \therefore x=\pm 2

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} x} -এর মান \fn_cm {\color{DarkGreen} \pm 2} হবে।

Koshe dekhi 9.3 class 10

Q6. যদি   \fn_cm {\color{Blue} a=\frac{\sqrt{5}+1}{\sqrt{5}-1}}  ও   \fn_cm {\color{Blue} b=\frac{\sqrt{5}-1}{\sqrt{5}+1}}  হয়, তবে নীচের মানগুলি নির্ণয় করি।

(i) \fn_cm {\color{Blue} \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}}

(ii) \fn_cm {\color{Blue} \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}}

(iii) \fn_cm {\color{Blue} \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}

(iv) \fn_cm {\color{Blue} \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}

সমাধানঃ 

\fn_cm a=\frac{\sqrt{5}+1}{\sqrt{5}-1} ,   \fn_cm b=\frac{\sqrt{5}-1}{\sqrt{5}+1}  

∴ a + b

\fn_cm =\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}

\fn_cm =\frac{\left ( \sqrt{5}+1 \right )^{2}+\left ( \sqrt{5}-1 \right )^{2}}{\left ( \sqrt{5}-1 \right )\left ( \sqrt{5}+1 \right )}

\fn_cm =\frac{2\left [ \left ( \sqrt{5} \right )^{2}+\left ( 1 \right )^{2} \right ]}{\left ( \sqrt{5} \right )^{2}-\left ( 1 \right )^{2}} \fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}

\fn_cm =\frac{2\left ( 5+1 \right )}{\left ( 5-1 \right )}

\fn_cm =\frac{2\times 6}{4}

\fn_cm =3

এবং,

\fn_cm \left ( a-b \right ) \fn_cm =\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}

\fn_cm =\frac{\left ( \sqrt{5}+1 \right )^{2}-\left ( \sqrt{5}-1 \right )^{2}}{\left ( \sqrt{5}-1 \right )\left ( \sqrt{5}+1 \right )}

\fn_cm =\frac{4\times \sqrt{5}\times 1}{\left ( \sqrt{5} \right )^{2}-\left ( 1 \right )^{2}} \fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}-\left ( a-b \right )^{2}=4ab, \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}

\fn_cm =\frac{4\sqrt{5}}{\left ( 5-1 \right )}

\fn_cm =\frac{4\sqrt{5}}{4}

\fn_cm =\sqrt{5}

আবার,

\fn_cm ab=\frac{\left (\sqrt{5}+1 \right )}{\left (\sqrt{5}-1 \right )}\times \frac{\left (\sqrt{5}-1 \right )}{\left (\sqrt{5}+1 \right )}

\fn_cm \therefore ab=1

 

সমাধান (i) :

এখন, (i) \fn_cm \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}

\fn_cm =\frac{\left \{ \left ( a+b \right )^{2}-2ab \right \}+ab}{\left \{ \left ( a+b \right )^{2}-2ab \right \}-ab} {\color{Blue} \left [ \because a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab \right ]}

\fn_cm =\frac{\left ( 3 \right )^{2}-2\times 1+1}{\left ( 3 \right )^{2}-2\times 1-1} [ \fn_cm {\color{Blue} a+b=3} এবং \fn_cm {\color{Blue} ab=1} বসিয়ে পাই ]

\fn_cm =\frac{9-2+1}{9-2-1}

\fn_cm =\frac{8}{6}

\fn_cm =\frac{4}{3}

\fn_cm =1\frac{1}{3}

উত্তরঃ নির্ণেয়  \fn_cm {\color{DarkGreen} \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}} এর মান \fn_cm {\color{DarkGreen} 1\frac{1}{3}}

 

 

সমাধান (ii) :

(ii) \fn_cm \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}

\fn_cm =\frac{\left ( \sqrt{5} \right )^{3}}{\left ( 3 \right )^{3}}   \fn_cm {\color{Blue}[\because a-b=\sqrt{5}} এবং \fn_cm {\color{Blue} a+b=3]} 

\fn_cm =\frac{5\sqrt{5}}{27}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}}  এর মান \fn_cm {\color{DarkGreen} \frac{5\sqrt{5}}{27}}

 

সমাধান (iii) :

(iii) \fn_cm \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}

\fn_cm =\frac{3\left ( a^{2} +b^{2} \right )+5ab}{3\left ( a^{2} +b^{2} \right )-5ab}

\fn_cm =\frac{3\left [\left ( a+b \right )^{2}-2ab \right ]+5ab}{3\left [\left ( a+b \right )^{2}-2ab \right ]-5ab} {\color{Blue} \left [ \because a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab \right ]}

\fn_cm =\frac{3\left [\left ( 3 \right )^{2}-2\times 1 \right ]+1}{3\left [\left ( 3 \right )^{2}-2\times 1 \right ]-1} [ \fn_cm {\color{Blue} a+b=3} এবং \fn_cm {\color{Blue} ab=1} বসিয়ে পাই ]

\fn_cm =\frac{3\left [ 9-2 \right ]+5\times 1}{3\left [ 9-2 \right ]-5\times 1}

\fn_cm =\frac{3\times 7+5}{3\times 7-5}

\fn_cm =\frac{26}{16}

\fn_cm =\frac{13}{8}

\fn_cm =1\frac{5}{8}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}  এর মান \fn_cm {\color{DarkGreen} 1\frac{5}{8}}

 

 

সমাধান (iv) :

(iv) \fn_cm \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}

\fn_cm =\frac{\left ( a+b \right )^{3}-3ab\left ( a+b \right )}{\left ( a-b \right )^{3}+3ab\left ( a-b \right )}

\fn_cm =\frac{\left ( 3 \right )^{3}-3\times 1\times 3}{\left ( \sqrt{5} \right )^{3}+3\times 1\times \sqrt{5}}

\fn_cm =\frac{27-9}{5\sqrt{5}+3\sqrt{5}}

\fn_cm =\frac{18}{8\sqrt{5}}

\fn_cm =\frac{9}{4\sqrt{5}}

\fn_cm =\frac{9\times \sqrt{5}}{4\sqrt{5}\times \sqrt{5}}

\fn_cm =\frac{9\sqrt{5}}{20}

উত্তরঃ নির্ণেয়  \fn_cm {\color{DarkGreen} \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}  এর মান \fn_cm {\color{DarkGreen} \frac{9\sqrt{5}}{20}}

Koshe dekhi 9.3 class 10

Q7. যদি  \fn_cm {\color{Blue} x=2+\sqrt{3}}  এবং  \fn_cm {\color{Blue} y=2-\sqrt{3}}  হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্ণয় করি :

(a)

(i) \fn_cm {\color{Blue} x-\frac{1}{x}}

(ii) \fn_cm {\color{Blue} {{y}^{2}}+\frac{1}{{{y}^{2}}}}

(iii) \fn_cm {\color{Blue} {{x}^{3}}-\frac{1}{{{x}^{3}}}}

(iv) \fn_cm {\color{Blue} xy+\frac{1}{xy}}

(b) \fn_cm {\color{Blue} 3{{x}^{2}}-5xy+3{{y}^{2}}}

সমাধানঃ 

\fn_cm \because x=2+\sqrt{3}

\fn_cm \therefore \frac{1}{x}=\frac{1}{2+\sqrt{3}}

বা, \fn_cm \frac{1}{x}=\frac{\left (2-\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )} [ হরের করণী নিরসন করে পাই ]

বা, \fn_cm \frac{1}{x}=\frac{2-\sqrt{3}}{\left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}} \fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

বা, \fn_cm \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}\Rightarrow 2-\sqrt{3}

\fn_cm \therefore \frac{1}{x}=2-\sqrt{3}

একইভাবে,

\fn_cm \because y=2-\sqrt{3}

\fn_cm \therefore \frac{1}{y}=\frac{1}{2-\sqrt{3}}

বা, \fn_cm \frac{1}{y}=\frac{\left (2+\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )} [ হরের করণী নিরসন করে পাই ]

বা, \fn_cm \frac{1}{y}=\frac{2+\sqrt{3}}{\left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}} \fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

বা, \fn_cm \frac{1}{y}=\frac{2+\sqrt{3}}{4-3}\Rightarrow 2+\sqrt{3}

\fn_cm \therefore \frac{1}{y}=2+\sqrt{3}

 

সমাধান (i) :

এখন, (i) \fn_cm {\color{Blue} x-\frac{1}{x}}

\fn_cm =\left ( 2+\sqrt{3} \right )-\left ( 2-\sqrt{3} \right ) [ \fn_cm {\color{Blue} x}\fn_cm {\color{Blue} \frac{1}{x}} -এর মান বসিয়ে পাই ]

\fn_cm =2+\sqrt{3}-2+\sqrt{3}

\fn_cm =2\sqrt{3}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} x-\frac{1}{x}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 2\sqrt{3}} .

 

সমাধান (ii) :

(ii) \fn_cm {\color{Blue} {{y}^{2}}+\frac{1}{{{y}^{2}}}}

\fn_cm \because y=2-\sqrt{3} এবং \fn_cm \frac{1}{y}=2+\sqrt{3}

\fn_cm \therefore y+\frac{1}{y}

\fn_cm =\left ( 2-\sqrt{3} \right )+\left ( 2+\sqrt{3} \right ) [ \fn_cm {\color{Blue} y}\fn_cm {\color{Blue} \frac{1}{y}} -এর মান বসিয়ে পাই ]

\fn_cm =2-\sqrt{3}+2+\sqrt{3}

\fn_cm =4

সুতরাং, \fn_cm y+\frac{1}{y}=4

এখন,  \fn_cm {{y}^{2}}+\frac{1}{{{y}^{2}}}

\fn_cm =\left ( y+\frac{1}{y} \right )^{2}-2\cdot y\cdot \frac{1}{y}

\fn_cm =4^{2}-2\; {\color{Blue} \left [\because y+\frac{1}{y}=4 \right ]}

\fn_cm =16-2\Rightarrow 14

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{y}^{2}}+\frac{1}{{{y}^{2}}}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 14}.

 

সমাধান (iii) :

(iii) \fn_cm {\color{Blue} {{x}^{3}}-\frac{1}{{{x}^{3}}}}

\fn_cm =\left ( x-\frac{1}{x} \right )^{3}+3\cdot x\cdot \frac{1}{x}\left (x-\frac{1}{x} \right )\; {\color{Blue} \left [ \because x-\frac{1}{x}=2\sqrt{3} \right ]}

\fn_cm =\left ( 2\sqrt{3} \right )^{3}+3\times 2\sqrt{3}

\fn_cm =24\sqrt{3}+6\sqrt{3}

\fn_cm =30\sqrt{3}

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} {{x}^{3}}-\frac{1}{{{x}^{3}}}} এর  সরলতম মান \fn_cm {\color{DarkGreen} 30\sqrt{3}}.

 

সমাধান (iv) :

(iv) \fn_cm {\color{Blue} xy+\frac{1}{xy}}

\fn_cm \because x=2+\sqrt{3} এবং  \fn_cm y=2-\sqrt{3} 

\fn_cm \therefore xy=\left ( 2+\sqrt{3} \right )\left ( 2-\sqrt{3} \right )

\fn_cm =\left ( 2 \right )^{2}-\left ( 2\sqrt{3} \right )^{2}

\fn_cm =4-3

\fn_cm =1

\fn_cm \therefore xy+\frac{1}{xy}

\fn_cm =1+\frac{1}{1} [ \fn_cm {\color{Blue} xy=1} বসিয়ে পাই ]

\fn_cm =1+1

\fn_cm =2

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} xy+\frac{1}{xy}} এর  সরলতম মান 2.

 

সমাধান (b) :

(b) \fn_cm {\color{Blue} 3{{x}^{2}}-5xy+3{{y}^{2}}}

\fn_cm \because x=2+\sqrt{3} এবং  \fn_cm y=2-\sqrt{3} 

\fn_cm \therefore x+y=\left ( 2+\sqrt{3} \right )+\left ( 2-\sqrt{3} \right )

\fn_cm =2+\sqrt{3}+2-\sqrt{3}

\fn_cm =4

এখন, \fn_cm 3x^{2}-5xy+3y^{2}

\fn_cm =3\left ( x^{2}+y^{2} \right )-5xy

\fn_cm =3\left \{ \left ( x+y \right )^{2}-2xy \right \}-5xy

\fn_cm =3\left \{ \left ( 4 \right )^{2}-2\times 1 \right \}-5\times 1 \fn_cm {\color{Blue} \left [ \because x+y=4,\; xy=1 \right ]}

\fn_cm =3\left ( 16-2 \right )-5

\fn_cm =3\times 14-5\Rightarrow 42-5

\fn_cm =37

উত্তরঃ নির্ণেয় \fn_cm {\color{DarkGreen} 3{{x}^{2}}-5xy+3{{y}^{2}}} এর  সরলতম মান \fn_cm {\color{DarkGreen}37}.

Koshe dekhi 9.3 class 10

Q8.  \small {\color{Blue} x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}}  এবং  \small {\color{Blue} xy=1}  হলে, দেখাই যে, \small {\color{Blue} \frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}=\frac{12}{11}}

সমাধানঃ 

প্রদত্ত,

\fn_cm \because x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} এবং  \fn_cm xy=1 

\fn_cm \therefore y=\frac{1}{x}

\fn_cm =\frac{1}{\left (\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \right )}

\fn_cm =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}

এবং, \fn_cm x+y

\fn_cm =\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}} [ x  ও  y  এর মান বসিয়ে পাই ]

\fn_cm =\frac{\left ( \sqrt{7}+\sqrt{3} \right )^{2}+\left ( \sqrt{7}-\sqrt{3} \right )^{2}}{\left ( \sqrt{7}+\sqrt{3} \right )\left ( \sqrt{7}-\sqrt{3} \right )}

\fn_cm =\frac{2\left [ \left ( \sqrt{7} \right )^{2}+\left ( \sqrt{3} \right )^{2} \right ]}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{3} \right )^{2}} \fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}

\fn_cm =\frac{2\left ( 7+3 \right )}{\left ( 7-3 \right )}

\fn_cm =\frac{20}{4}=5

\fn_cm \therefore x+y=5

বামপক্ষ:

\fn_cm \frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}

\fn_cm =\frac{\left ( x^{2}+y^{2} \right )+xy}{\left ( x^{2}-y^{2} \right )-xy}

\fn_cm =\frac{\left [ \left ( x+y \right )^{2}-2xy \right ]+xy}{\left [ \left ( x+y \right )^{2}-2xy \right ]-xy}

\fn_cm =\frac{\left ( 5 \right )^{2}-2\times 1+1}{\left ( 5 \right )^{2}-2\times 1-1} \fn_cm {\color{Blue} \left [ \because x+y=5,xy=1 \right ]}

\fn_cm =\frac{25-2+1}{25-2-1}

\fn_cm =\frac{24}{22}

\fn_cm =\frac{12}{11}= ডানপক্ষ ( প্রমাণিত )

Koshe dekhi 9.3 class 10

Q9.  \fn_cm {\color{Blue} \left( \sqrt{7}+1 \right)}  এবং  \fn_cm {\color{Blue} \left( \sqrt{5}+\sqrt{3} \right)}  এর মধ্যে কোনটি বড়ো লিখি।

সমাধানঃ 

\fn_jvn \left( \sqrt{7}+1 \right)^{2} [ বর্গ করে পাই ]

\fn_jvn =7+2\sqrt{7}+1

\fn_jvn =8+2\sqrt{7}

আবার,

  \fn_jvn \left( \sqrt{5}+\sqrt{3} \right)^{2}  [ বর্গ করে পাই ]

\fn_jvn =5+2\sqrt{15}+3

\fn_jvn =8+2\sqrt{15}

\fn_jvn \because \sqrt{15}>\sqrt{7}

\fn_jvn \therefore \left ( \sqrt{5}+\sqrt{3} \right )^{2}>\left( \sqrt{7}+1 \right)^{2}

বা, \fn_jvn \left ( \sqrt{5}+\sqrt{3} \right )>\left( \sqrt{7}+1 \right)

উত্তরঃ নির্ণেয় \fn_jvn {\color{DarkGreen} \left( \sqrt{7}+1 \right)}  এবং  \fn_jvn {\color{DarkGreen} \left( \sqrt{5}+\sqrt{3} \right)}  এর মধ্যে \fn_jvn {\color{DarkGreen} \left( \sqrt{5}+\sqrt{3} \right)}  বড়ো। 

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(i)  \small {\color{Blue} x=2+\sqrt{3}}   হলে,  \small {\color{Blue} x+\frac{1}{x}}  এর মান –

(a) 2

(b) 2√3

(c) 4

(d) 2 − √3

সমাধানঃ 

\fn_jvn x=2+\sqrt{3} ,

\fn_jvn \frac{1}{x}=\frac{1}{2+\sqrt{3}}

বা, \fn_jvn \frac{1}{x}=\frac{\left ( (2-\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )} [ হরের করণী নিরসন করে পাই ]

বা, \fn_jvn \frac{1}{x}=\frac{2-\sqrt{3}}{\left [ \left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}\right ]} \fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_jvn \therefore \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}

\fn_jvn \therefore x+\frac{1}{x}

\fn_jvn =2+\sqrt{3}+2-\sqrt{3}

\fn_jvn =4

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ( c \right )4}

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(ii) যদি  \small {\color{Blue} p+q=\sqrt{13}}  এবং  \small {\color{Blue} p-q=\sqrt{5}}  হয়, তাহলে \small {\color{Blue} pq} এর মান –

(a) 2

(b) 18

(c) 9

(d) 8

সমাধানঃ 

  \fn_jvn p+q=\sqrt{13}

বা, \fn_jvn \left ( p+q \right )^{2}=\left ( \sqrt{13} \right )^{2} [ উভয়পক্ষে বর্গ করে পাই ]

বা, \fn_jvn \left ( p+q \right )^{2}=13......\left ( i \right )

আবার,

  \fn_jvn p-q=\sqrt{5} 

বা, \fn_jvn \left ( p-q \right )^{2}=\left ( \sqrt{5} \right )^{2} [ উভয়পক্ষে বর্গ করে পাই ]

বা, \fn_jvn \left ( p-q \right )^{2}=5......\left ( ii \right )

\fn_jvn \left ( i \right ) নং সমীকরণ থেকে \fn_jvn \left ( ii \right ) নং সমীকরণ বিয়োগ করে পাই 

\fn_jvn \left ( p+q \right )^{2}-\left ( p-q \right )^{2}=13-5

বা, \fn_jvn 4pq=8 {\color{Blue} \left [ \because \left ( a+b \right )^{2}-\left ( a-b \right )^{2}=4ab \right ]}

\fn_jvn \therefore pq=\frac{8}{4}=2

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ( a \right )2}

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(iii) যদি   \small {\color{Blue} a+b=\sqrt{5}}  এবং   \small {\color{Blue} a-b=\sqrt{3}}  তাহলে   \small {\color{Blue} \left( {{a}^{2}}+{{b}^{2}} \right)}  এর মান –

(a) 8

(b) 4

(c) 2

(d) 1

সমাধানঃ 

\fn_jvn a+b=\sqrt{5} 

বা, \fn_jvn \left ( a+b \right )^{2}=\left ( \sqrt{5} \right )^{2} [ উভয়পক্ষে বর্গ করে পাই ]

বা, \fn_jvn \left ( a+b \right )^{2}=5......\left ( i \right )

আবার,

\fn_jvn a-b=\sqrt{3} 

বা, \fn_jvn \left ( a-b \right )^{2}=\left ( \sqrt{3} \right )^{2} [ উভয়পক্ষে বর্গ করে পাই ]

বা, \fn_jvn \left ( a-b \right )^{2}=3......\left ( ii \right )

\fn_jvn \left ( i \right ) নং সমীকরণ ও \fn_jvn \left ( ii \right ) নং সমীকরণ যোগ করে পাই 

\fn_jvn \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=5+3

বা, \fn_jvn a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}=8

বা, \fn_jvn 2\left( {{a}^{2}}+{{b}^{2}} \right)=8

\fn_jvn \therefore \left( {{a}^{2}}+{{b}^{2}} \right)=\frac{8}{2}=4  

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ( b \right )4}

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(iv)  √125  থেকে  √5  বিয়োগ করলে বিয়োগফল হবে –

(a) √80

(b) √120

(c) √100

(d) কোনটিই নয়

সমাধানঃ 

\fn_jvn \sqrt{125}  থেকে  \fn_jvn \sqrt{5}  বিয়োগ করলে বিয়োগফল হবে-

\fn_jvn \sqrt{125}-\sqrt{5}

\fn_jvn =\sqrt{5\times 5\times 5}-\sqrt{5}

\fn_jvn =5\sqrt{5}-\sqrt{5}

\fn_jvn =4\sqrt{5}

\fn_jvn =\sqrt{4\times 4\times 5}

\fn_jvn =\sqrt{80}

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ( a \right )\sqrt{80}}

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(v)  \small {\color{Blue} \left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right)}  এর গুণফল –

(a) 22

(b) 44

(c) 2

(d) 11

সমাধানঃ 

\fn_jvn \left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right) এর গুণফল –

\fn_jvn \left( 5-\sqrt{3} \right)\times \left( \sqrt{3}-1 \right)\times \left( 5+\sqrt{3} \right)\times \left( \sqrt{3}+1 \right)

\fn_jvn =\left [\left( 5-\sqrt{3} \right)\times \left( 5+\sqrt{3} \right) \right ]\times\left [ \left( \sqrt{3}-1 \right)\times \left( \sqrt{3}+1 \right) \right ]

\fn_jvn =\left [\left (5 \right )^{2}-\left (\sqrt{3} \right )^{2} \right ]\times\left [ \left( \sqrt{3}\right)^{2}-\left ( 1 \right )^{2} \right ] {\color{Blue} \left [ \because \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2} \right ]}

\fn_jvn =\left [ 25-3 \right ]\times \left [ 3-1 \right ]

\fn_jvn =22\times 2

\fn_jvn =44

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ( b \right )44}

Koshe dekhi 9.3 class 10

(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি :

(i)   \fn_jvn \sqrt{75}  এবং  \fn_jvn \sqrt{147}  সদৃশ করণী।

সমাধানঃ 

  \fn_jvn \sqrt{75} 

\fn_jvn =\sqrt{5\times 5\times 3}

\fn_jvn =5\sqrt{3}

এবং 

\fn_jvn \sqrt{147}

\fn_jvn =\sqrt{7\times 7\times 3}

\fn_jvn =7\sqrt{3}

∴  \fn_jvn \sqrt{75}  এবং  \fn_jvn \sqrt{147}  সদৃশ করণী।

উত্তরঃ বিবৃতিটি সত্য। 

 

(ii)  \fn_jvn \sqrt{\pi}  একটি দ্বিঘাত করণী।

সমাধানঃ 

বিবৃতিটি মিথ্যা। 

কারণ : যে সমস্ত সংখ্যাকে  \fn_jvn \pm \sqrt{a} আকারে লেখা যায়, যেখানে  \fn_jvn a  হলো ধনাত্মক পূর্ণসংখ্যা। 

যেহেতু, \fn_jvn \pi পূর্ণসংখ্যা নয় (\fn_jvn \pi হলো অমূলদ সংখ্যা ), তাই \fn_jvn \sqrt{\pi}  দ্বিঘাত করোনি নয়। 

উত্তরঃ বিবৃতিটি মিথ্যা।

 

(C) শূন্যস্থান পূরণ করি :

(i)  \fn_jvn 5\sqrt{11}  একটি _______ সংখ্যা। (মূলদ / অমূলদ)

উত্তরঃ অমূলদ [ যেহেতু \fn_jvn {\color{Blue} 5\sqrt{11}} সংখ্যাটির সুনির্দিষ্ট দশমিক মান সম্পূর্ণভাবে নির্ণয় করা যায় না, তাই এটি একটি অমূলদ সংখ্যা ]

 

(ii)  \fn_jvn \left ( \sqrt{3}-5 \right )  এর অনুবন্ধী করণী _______ ।

সমাধানঃ 

\fn_jvn \left ( \sqrt{3}-5 \right )  ও \fn_jvn \left (-\sqrt{3}-5 \right ) এর যোগফল  

  \fn_jvn \left ( \sqrt{3}-5 \right )+\left (- \sqrt{3}-5 \right )

\fn_jvn =-10 ( মূলদ সংখ্যা )  

\fn_jvn \left ( \sqrt{3}-5 \right )  ও \fn_jvn \left (-\sqrt{3}-5 \right ) এর গুনফল 

\fn_jvn \left ( \sqrt{3}-5 \right )\times \left (- \sqrt{3}-5 \right )

\fn_jvn =-3-5\sqrt{3}+5\sqrt{3}+25

\fn_jvn =-3+25

\fn_jvn =22 ( মূলদ সংখ্যা )  

উত্তরঃ \fn_jvn {\color{DarkGreen} \left ({\color{DarkGreen} -\sqrt{3}-5} \right )} [ যেহেতু \fn_jvn {\color{Blue} \left ( \sqrt{3}-5 \right )}  ও \fn_jvn {\color{Blue} \left (-\sqrt{3}-5 \right )} এর যোগফল এবং গুনফল উভয়ই মূলদ সংখ্যা ]

 

(iii) দুটি দ্বিঘাত করণীর যোগফল ও গুনফল একটি মূলদ সংখ্যা হলে, করণীদ্বয় ________ করণী।

উত্তরঃ অনুবন্ধী

 

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(i)  \small {\color{Blue} x=3+2\sqrt{2}}  হলে,  \small {\color{Blue} x+\frac{1}{x}}  -এর মান লিখি।

সমাধানঃ 

\fn_jvn x=3+2\sqrt{2}

\fn_jvn \therefore \frac{1}{x}=\frac{1}{3+2\sqrt{2}}

\fn_jvn =\frac{\left ( (3-2\sqrt{2} \right )}{\left (3+2\sqrt{2} \right )\left ( (3-2\sqrt{2} \right )} [ হরের করণী নিরসন করে পাই ]

\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{\left [\left (3 \right )^{2} \right ]-\left [\left (2\sqrt{2} \right )^{2} \right ] } \fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{9-8}

\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{1}

\fn_jvn =\left ( 3-2\sqrt{2} \right )

\fn_jvn \therefore x+\frac{1}{x}

\fn_jvn =3+2\sqrt{2}+3-2\sqrt{2} {\color{Blue} \left [ \because x=3+2\sqrt{2},\frac{1}{x}=3-2\sqrt{2} \right ]}

\fn_jvn =6

উত্তরঃ নির্ণেয় \fn_jvn {\color{DarkGreen} \left (x+\frac{1}{x} \right )} এর মান \fn_jvn {\color{DarkGreen} 6} .

 

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(ii)  \small {\color{Blue} \left ( \sqrt{15}+\sqrt{3} \right )}  এবং  \small {\color{Blue} \left ( \sqrt{10}+\sqrt{8} \right )}  এর মধ্যে কোনটি বড় লিখি।

সমাধানঃ 

\fn_jvn \left ( \sqrt{15}+\sqrt{3} \right )^{2} [ বর্গ করে পাই ]

\fn_jvn =15+2\sqrt{45}+3

\fn_jvn =18+2\sqrt{3\times 3\times 5}

\fn_jvn =18+6\sqrt{5}

আবার,

  \fn_jvn \left ( \sqrt{10}+\sqrt{8} \right )^{2}  [ বর্গ করে পাই ]

\fn_jvn =10+2\sqrt{80}+8

\fn_jvn =18+2\sqrt{5\times 4\times 4}

\fn_jvn =18+8\sqrt{5}

\fn_jvn \because 8\sqrt{5}>6\sqrt{5}

\fn_jvn \therefore \left ( \sqrt{10}+\sqrt{8} \right )^{2}>\left( \sqrt{15}+\sqrt{3} \right)^{2}

বা, \fn_jvn \left ( \sqrt{10}+\sqrt{8} \right )>\left( \sqrt{15}+\sqrt{3} \right)

উত্তরঃ নির্ণেয় \fn_jvn {\color{DarkGreen} \left ( \sqrt{15}+\sqrt{3} \right )}  এবং  \fn_jvn {\color{DarkGreen} \left ( \sqrt{10}+\sqrt{8} \right )} এর মধ্যে \fn_jvn {\color{DarkGreen} \left ( \sqrt{10}+\sqrt{8} \right )}  বড়ো। 

 

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(iii) দুটি মিশ্র দ্বিঘাত করণী লিখি যাদের গুণফল একটি মূলদ সংখ্যা।

সমাধানঃ 

যদি দুটি মিশ্র দ্বিঘাত করণী সদৃশ্য হয়, তবে তাদের গুনফল হবে।

যেমন,  \fn_jvn 2\sqrt{3}\fn_jvn 5\sqrt{3} হলো দুটি মিশ্র সদৃশ্য দ্বিঘাত করণী যাদের গুনফল নিম্নরুপঃ 

\fn_jvn 2\sqrt{3}\times 5\sqrt{3}

\fn_jvn =\left ( 2\times 5 \right )\times \left ( \sqrt{3}\times \sqrt{3} \right )

\fn_jvn =10\times \left ( 3 \right )\; \; \; {\color{Blue} \left [ \because \sqrt{a}\times \sqrt{a}=a \right ]}

\fn_jvn = 30 … এটি হলো একটি মূলদ সংখ্যা। 

উত্তরঃ নির্ণেয় মিশ্র করণী দুটি হলো \fn_jvn {\color{DarkGreen} 2\sqrt{3}}\fn_jvn {\color{DarkGreen} 5\sqrt{3}}

 

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(iv)  √72  থেকে কত বিয়োগ করলে √32  হবে তা লিখি।

সমাধানঃ 

ধরি,   \fn_jvn \sqrt{72} থেকে \fn_jvn x বিয়োগ করলে \fn_jvn \sqrt{32}  হবে। 

\fn_jvn \therefore \sqrt{72}-x=\sqrt{32}

বা, \fn_jvn 6\sqrt{2}-x=4\sqrt{2}

বা, \fn_jvn 6\sqrt{2}-4\sqrt{2}=x

\fn_jvn \therefore x=2\sqrt{2}

উত্তরঃ নির্ণেয় \fn_jvn {\color{DarkGreen} \sqrt{72}} থেকে \fn_jvn {\color{DarkGreen} 2\sqrt{2}} বিয়োগ করলে \fn_jvn {\color{DarkGreen} \sqrt{32}}  হবে। 

 

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(v)  \small {\color{Blue} \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]} – এর সরলতম মান লিখি।

সমাধানঃ 

  \fn_jvn \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]

\fn_jvn =\left[ \frac{\left (\sqrt{2}-1 \right )}{\left (\sqrt{2}+1 \right )\left (\sqrt{2}-1 \right )}+\frac{\left (\sqrt{3}-\sqrt{2} \right )}{\left (\sqrt{3}+\sqrt{2} \right )\left (\sqrt{3}-\sqrt{2} \right )}+\frac{\left (\sqrt{4}-\sqrt{3} \right )}{\left (\sqrt{4}+\sqrt{3} \right )\left (\sqrt{4}-\sqrt{3} \right )} \right] [ হরের করণী নিরসন করে পাই ]

\fn_jvn =\frac{\left ( \sqrt{2}-1 \right )}{\left [\left (\sqrt{2} \right )^{2} \right ]-\left [\left (1 \right )^{2} \right ] }+\frac{\left ( \sqrt{3}-\sqrt{2} \right )}{\left [\left (\sqrt{3} \right )^{2} \right ]-\left [\left (\sqrt{2}\right )^{2} \right ] }+\frac{\left ( \sqrt{4}-\sqrt{3} \right )}{\left [\left (\sqrt{4} \right )^{2} \right ]-\left [\left (\sqrt{3}\right )^{2} \right ] } \fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}

\fn_jvn =\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}

\fn_jvn =\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}

\fn_jvn =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}

\fn_jvn =\sqrt{4}-1

\fn_jvn =2-1

\fn_jvn =1

উত্তরঃ নির্ণেয়   \fn_jvn {\color{DarkGreen} \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]} – এর সরলতম মান \fn_jvn {\color{DarkGreen} 1} .

Koshe dekhi 9.3 class 10

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