Koshe dekhi 15 Class 8

Table of Contents

Koshe dekhi 15 Class 8

1. নীচের সম্পর্কগুলি দেখি ও কোনটি সত্য ও কোনটি মিথ্যা লিখি।

(i) \(=\frac{a+b}{c} =\frac{a}{c}+\frac{b}{c}\)

সমাধানঃ
বামপক্ষ :
\(=\frac{a+b}{c}\)

ডানপক্ষ :
\( =\frac{a}{c}+\frac{b}{c} \)
\( =\frac{a+b}{c}\)

\(\because \) বামপক্ষ = ডানপক্ষ
\( \therefore \) প্রদত্ত সম্পর্কটি সত্য।

(ii) \( \frac{a}{x+y} =\frac{a}{x}+\frac{b}{y} \)

সমাধানঃ
বামপক্ষ:
\(
=\frac{a}{x+y}
\)

ডানপক্ষ:
\( =\frac{a}{x}+\frac{b}{y} \)
\( =\frac{a y+b x}{x y} \neq \) ডানপক্ষ

\(\because \) বামপক্ষ ≠ ডানপক্ষ
\( \therefore \) প্রদত্ত সম্পর্কটি মিথ্যা।

(iii) \( \frac{x-y}{a-b}=\frac{y-x}{b-a} \)

সমাধানঃ
বামপক্ষ:
\( \frac{x-y}{a-b} \)
\( =\frac{-(y-x)}{-(b-a)}\)
\( =\frac{y-x}{b-a} =\) ডানপক্ষ

\(\because \) বামপক্ষ = ডানপক্ষ
\( \therefore \) প্রদত্ত সম্পর্কটি সত্য।

(iv) \( \frac{1}{x}+\frac{1}{y}=\frac{1}{x+y} \)

সমাধানঃ
বামপক্ষ:
\(\frac{1}{x}+\frac{1}{y} \)
\( =\frac{y+x}{x y} \neq \) ডানপক্ষ

\(\because \) বামপক্ষ ≠ ডানপক্ষ
\( \therefore \) প্রদত্ত সম্পর্কটি মিথ্যা।

আরও দেখুনঃ

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান।

জীবন বিজ্ঞান (দশম শ্রেণী) (Life Science)

2. নীচের বীজগাণিতিক ভগ্নাংশগুলি লঘিষ্ঠ আকারে প্রকাশ করি।

(i) \(\frac{{63{{a}^{3}}{{b}^{4}}}}{{77{{b}^{5}}}}\)

সমাধানঃ

\(\frac{{63{{a}^{3}}{{b}^{4}}}}{{77{{b}^{5}}}}\)
\( =\frac{{9{{a}^{3}}}}{{11{{b}^{{5-4}}}}}\quad \left[ {\because \frac{{{{x}^{m}}}}{{{{x}^{n}}}}=\frac{1}{{{{x}^{{n-m}}}}};\quad n>m} \right]\)
\( =\frac{{9{{a}^{3}}}}{{11b}}\)

(ii) \( \frac{{18{{a}^{4}}{{b}^{5}}{{c}^{2}}}}{{21{{a}^{7}}{{b}^{2}}}} \)

সমাধানঃ

\(\frac{{18{{a}^{4}}{{b}^{5}}{{c}^{2}}}}{{21{{a}^{7}}{{b}^{2}}}}\)
\(=\frac{{6{{b}^{{5-2}}}{{c}^{2}}}}{{7{{a}^{{7-4}}}}}\quad \left[ \begin{array}{l}\because \frac{{{{x}^{m}}}}{{{{x}^{n}}}}={{x}^{{m-n}}};\quad m>n\\\frac{{{{x}^{m}}}}{{{{x}^{n}}}}=\frac{1}{{{{x}^{{n-m}}}}};\quad n>m\end{array} \right]\)
\(=\frac{{6{{b}^{3}}{{c}^{2}}}}{{7{{a}^{3}}}}\)

(iii) \( \frac{x^{2}-3 x+2}{x^{2}-1} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{x^{2}-3 x+2}{x^{2}-1} \\ =\frac{x^{2}-(2+1)x+2}{x^{2}-1} \\
=\frac{x^{2}-2 x-x+2}{x^{2}-1^{2}} \\
=\frac{x(x-2)-1(x-2)}{(x+1)(x-1)} \\
=\frac{(x-2)(x-1)}{(x+1)(x-1)} \\ =\frac{x-2}{x+1}
\end{array}
\)

(iv) \( \frac{a+1}{a-2} \times \frac{a^{2}-a-2}{a^{2}+a} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{a+1}{a-2} \times \frac{a^{2}-a-2}{a^{2}+a} \\ =\frac{a+1}{a-2} \times \frac{a^{2}-(2-1)a-2}{a(a+1)} \\
=\frac{(a+1)}{(a-2)} \times \frac{a^{2}-2 a+a-2}{a(a+1)} \\
=\frac{(a+1)}{(a-2)} \times \frac{a(a-2)+1(a-2)}{a(a+1)} \\
=\frac{1}{(a-2)} \times \frac{(a-2)(a+1)}{a} \\
=\frac{a+1}{a}
\end{array}
\)

(v) \( \frac{p^{3}+p^{3}}{p^{2}-q^{2}} \div \frac{p+q}{p-q} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{p^{3}+p^{3}}{p^{2}-q^{2}} \div \frac{p+q}{p-q} \\
=\frac{p^{3}+q^{3}}{p^{2}-q^{2}} \times \frac{(p-q)}{(p+q)} \\
=\frac{(p+q)\left(p^{2}+p q+q^{2}\right)}{(p+q)(p-q)} \times \frac{(p-q)}{(p+q)} \\
=\frac{p^{2}-p q+q^{2}}{p+q}
\end{array}
\)

(vi) \( \frac{x^{2}-x-6}{x^{2}+4 x-5} \times \frac{x^{2}+6 x+5}{x^{2}-4 x+3} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{x^{2}-x-6}{x^{2}+4 x-5} \times \frac{x^{2}+6 x+5}{x^{2}-4 x+3} \\
=\frac{x^{2}-(3-2)x-6}{x^{2}+(5-1) x-5} \times \frac{x^{2}+(5+1) x+5}{x^{2}-(3+1)x+3} \\
=\frac{x^{2}-3 x+2 x-6}{x^{2}+5 x-x-5} \times \frac{x^{2}+5 x+x+5}{x^{2}-3 x-x+3} \\
=\frac{x(x-3)+2(x-3)}{x(x+5)-1(x+5)} \times \frac{x(x+5)+1(x+5)}{x(x-3)-1(x-3)} \\
=\frac{(x-3)(x+2)}{(x+5)(x-1)} \times \frac{(x+5)(x+1)}{(x-1)(x-3)} \\
=\frac{x+2}{x-1} \times \frac{x+1}{x-1} \\
=\frac{x^{2}+2 x+x+2}{x^{2}-2 x+1} \\
=\frac{x^{2}+3 x+2}{x^{2}-2 x+1}
\end{array}
\)

(vii) \( \frac{a^{2}-a b+b^{2}}{a^{2}+a b} \div \frac{a^{3}+b^{3}}{a^{2}-b^{2}} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{a^{2}-a b+b^{2}}{a^{2}+a b} \div \frac{a^{3}+b^{3}}{a^{2}-b^{2}} \\
=\frac{\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b} \times \frac{a^{2}-b^{2}}{a^{3}+b^{3}} \\
=\frac{\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b} \times \frac{(a+b)(a-b)}{(a+b)\left(a^{2}-a b+b^{2}\right)} \\
=\frac{(a^{2}-a b+b^{2})}{a(a+b)} \times \frac{(a+b)(a-b)}{(a+b)(a^{2}-a b+b^{2})} \\
=\frac{(a-b)}{a(a+b)} \\ =\frac{(a-b)}{\left(a^{2}-a b\right)}
\end{array}
\)

আরও দেখুনঃ

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান।

জীবন বিজ্ঞান (দশম শ্রেণী) (Life Science)

3. নীচের বীজগাণিতিক ভগ্নাংশগুলি সরলতম আকারে প্রকাশ করি।

(i) \( \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \\
=\frac{a+b+c}{a b c}
\end{array}
\)

(ii) \( \frac{a-b-c}{a}+\frac{a+b+c}{a} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{a-b-c}{a}+\frac{a+b+c}{a} \\
=\frac{a-b-c+a+b+c}{a} \\
=\frac{2 a}{a} \\ =2
\end{array}
\)

(iii) \( \frac{x^{2}+a^{2}}{a b}+\frac{x-a}{a x}-\frac{x^{3}}{b} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{x^{2}+a^{2}}{a b}+\frac{x-a}{a x}-\frac{x^{3}}{b} \\
=\frac{x\left(x^{2}+a^{2}\right)+b(x-a)-a x \cdot x^{3}}{a b x} \\
=\frac{x^{3}+a^{2} x+b x-a b-a x^{4}}{a b x}
\end{array}
\)

(iv) \( \frac{2 a^{2} b}{3 b^{2} c} \times \frac{c^{4}}{3 a^{3}} \div \frac{4 b c^{3}}{9 a^{2}} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{2 a^{2} b}{3 b^{2} c} \times \frac{c^{4}}{3 a^{3}} \div \frac{4 b c^{3}}{9 a^{2}} \\
=\frac{2 a^{2} b}{3 b^{2} c} \times \frac{c^{4}}{3 a^{3}} \times \frac{9 a^{2}}{4 b c^{3}} \\
=\frac{(a)^{2+2-3}(c)^{4-1-3}}{2 (b)^{2+1-1}} \\
=\frac{a}{2 b^{2}}
\end{array}
\)

(v) \( \frac{1}{{x}^{2}-3x+2} + \frac{1}{{x}^{2}-5x+6} + \frac{1}{{x}^{2}-4x+3} \)

সমাধানঃ
\(
\begin{array}{l}{{x}^{2}}-3x+2\\={{x}^{2}}-\left( {2+1} \right)x+2\\={{x}^{2}}-2x-x+2\\=x\left( {x-2} \right)-1\left( {x-2} \right)\\=\left( {x-2} \right)\left( {x-1} \right)
\end{array}
\)

\( \begin{array}{l}{{x}^{2}}-5x+6\\={{x}^{2}}-\left( {3+2} \right)x+6\\={{x}^{2}}-3x-2x+6\\=x\left( {x-3} \right)-2\left( {x-3} \right)\\=\left( {x-3} \right)\left( {x-2} \right)\end{array} \)

\(\begin{array}{l}{{x}^{2}}-4x+3\\={{x}^{2}}-\left( {3+1} \right)x+3\\={{x}^{2}}-3x-x+3\\=x\left( {x-3} \right)-1\left( {x-3} \right)\\=\left( {x-3} \right)\left( {x-1} \right)\end{array}\)

এখন, \( \frac{1}{{x}^{2}-3x+2} + \frac{1}{{x}^{2}-5x+6} + \frac{1}{{x}^{2}-4x+3} \)

\( \begin{array}{l}=\frac{1}{{\left( {x-2} \right)\left( {x-1} \right)}}+\frac{1}{{\left( {x-3} \right)\left( {x-2} \right)}}+\frac{1}{{\left( {x-3} \right)\left( {x-1} \right)}}\\=\frac{{\left( {x-3} \right)+\left( {x-1} \right)+\left( {x-2} \right)}}{{\left( {x-3} \right)\left( {x-2} \right)\left( {x-1} \right)}}\\=\frac{{x-3+x-1+x-2}}{{\left( {x-3} \right)\left( {x-2} \right)\left( {x-1} \right)}}\\=\frac{{3x-6}}{{\left( {x-3} \right)\left( {x-2} \right)\left( {x-1} \right)}}\\=\frac{{3\left( {x-2} \right)}}{{\left( {x-3} \right)\left( {x-2} \right)\left( {x-1} \right)}}\\=\frac{3}{{\left( {x-3} \right)\left( {x-1} \right)}}\\=\frac{3}{x^{2}-x-3x+3} \\ =\frac{3}{x^{2}-4x+3}\end{array} \)

(vi) \( \frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^{2}+1}+\frac{4 x^{3}}{x^{4}+1} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^{2}+1}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{x+1+x-1}{(x-1)(x+1)}+\frac{2 x}{x^{2}+1}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{2 x}{x^{2}-1}+\frac{2 x}{x^{2}+1}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{2 x\left(x^{2}+1\right)+2 x\left(x^{2}-1\right)}{\left(x^{2}-1\right)\left(x^{2}+1\right)}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{2 x^{3}+2 x+2 x^{3}-2 x}{\left(x^{2}-1\right)\left(x^{2}+1\right)}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{4 x^{3}}{\left(x^{4}-1\right)}+\frac{4 x^{3}}{x^{4}+1} \\
=\frac{4 x^{3}\left(x^{4}+1\right)+4 x^{3}\left(x^{4}-1\right)}{\left(x^{4}-1\right)\left(x^{4}+1\right)} \\
=\frac{4 x^{7}+4 x^{3}+4 x^{7}-4 x^{3}}{\left(x^{4}-1\right)\left(x^{4}+1\right)}\\=\frac{8 x^{7}}{x^{8}-1}
\end{array}
\)

(vii) \( \frac{b^{2}-5 b}{3 b-4 a} \times \frac{9 b^{2}-16 a^{2}}{b^{2}-25} \div \frac{3 b^{2}+4 a b}{a b+5 a} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{b^{2}-5 b}{3 b-4 a} \times \frac{9 b^{2}-16 a^{2}}{b^{2}-25} \div \frac{3 b^{2}+4 a b}{a b+5 a} \\
=\frac{b^{2}-5 b}{3 b-4 a} \times \frac{9 b^{2}-16 a^{2}}{b^{2}-25} \times \frac{a b+5 a}{3 b^{2}+4 a b} \\
=\frac{b(b-5)}{3 b-4 a} \times \frac{(3 b)^{2}-(4 a)^{2}}{b^{2}-5^{2}} \times \frac{a(b+5)}{b(3 b+4 a)} \\
=\frac{b(b-5)}{3 b-4 a} \times \frac{(3 b+4 a)(3 b-4 a)}{(b+5)(b-5)} \times \frac{a(b+5)}{b(3 b+4 a)} \\
=a
\end{array}
\)

(viii) \( \frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c+b)} \\
=\frac{b+c}{(a-b)(a-c)}-\frac{c+a}{(a-b)(b-c)}+\frac{a+b}{(a-c)(b-c)} \\
=\frac{(b+c)(b-c)-(a+c)(a-c)+(a+b)(a-b)}{(a-b)(a-c)(b-c)} \\
=\frac{b^{2}-c^{2}-\left(a^{2}-c^{2}\right)+\left(a^{2}-b^{2}\right)}{(a-b)(a-c)(b-c)} \\
=\frac{b^{2}-c^{2}-a^{2}+c^{2}+a^{2}-b^{2}}{(a-b)(a-c)(b-c)} \\
=\frac{0}{(a-b)(a-c)(b-c)}=0
\end{array}
\)

আরও দেখুনঃ

গণিত প্রকাশ দশম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রকাশ নবম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা অষ্টম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা সপ্তম শ্রেণি – সম্পূর্ণ সমাধান।

গণিত প্রভা ষষ্ঠ শ্রেণি – সম্পূর্ণ সমাধান।

জীবন বিজ্ঞান (দশম শ্রেণী) (Life Science)

(ix) \( \frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)} \)

সমাধানঃ
\(
\begin{array}{l}
\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)} \\
= -\frac{b+c-a}{(a-b)(c-a)}-\frac{c+a-b}{(b-c)(a-b)}-\frac{a+b-c}{(c-a)(b-c)} \\
= -\left[\frac{b+c-a}{(a-b)(c-a)}+\frac{c+a-b}{(b-c)(a-b)}+\frac{a+b-c}{(c-a)(b-c)}\right] \\
= -\left[\frac{(b+c-a)(b-c)+(c+a-b)(c-a)+(a+b-c)(a-b)}{(a-b)(c-a)(b-c)}\right] \\ =-\left[ {\frac{{{{b}^{2}}-bc+bc-{{c}^{2}}-ab+{{c}^{2}}-ac+ac-{{a}^{2}}-bc+ab+{{a}^{2}}-ab+ab-{{b}^{2}}-ac+bc}}{{\left( {a-b} \right)\left( {c-a} \right)\left( {b-c} \right)}}} \right]\\
= -\left[\frac{0}{(a-b)(c-a)(b-c)}\right] \\
= 0
\end{array}
\)

(x) \( \frac{\frac{a^{2}}{x-a}+\frac{b^{2}}{x-b}+\frac{c^{2}}{x-b}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \)

সমাধানঃ
\(
\begin{array}{l}
=\frac{\frac{a^{2}}{x-a}+\frac{b^{2}}{x-b}+\frac{c^{2}}{x-b}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \\
=\frac{\frac{a^{2}}{x-a}+a+\frac{b^{2}}{x-b}+b+\frac{c^{2}}{x-b}+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \\
=\frac{\frac{a^{2}+a x-a^{2}}{x-a}+\frac{b^{2}+b x-b^{2}}{x-b}+\frac{c^{2}+c x-c^{2}}{x-b}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \\
=\frac{\frac{a x}{x-a}+\frac{b x}{x-b}+\frac{a}{x-b}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \\
=\frac{x\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-b}\right)}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}} \\
=x
\end{array}
\)

(xi) \( \left( {\frac{{{{a}^{2}}+{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}-\frac{{{{a}^{2}}-{{b}^{2}}}}{{{{a}^{2}}+{{b}^{2}}}}} \right)\div \left( {\frac{{a+b}}{{a-b}}-\frac{{a-b}}{{a+b}}} \right)\times \left( {\frac{a}{b}+\frac{b}{a}} \right) \)

সমাধানঃ

\( \begin{array}{l}\left( {\frac{{{{a}^{2}}+{{b}^{2}}}}{{{{a}^{2}}-{{b}^{2}}}}-\frac{{{{a}^{2}}-{{b}^{2}}}}{{{{a}^{2}}+{{b}^{2}}}}} \right)\div \left( {\frac{{a+b}}{{a-b}}-\frac{{a-b}}{{a+b}}} \right)\times \left( {\frac{a}{b}+\frac{b}{a}} \right)\\=\left[ {\frac{{{{{\left( {{{a}^{2}}+{{b}^{2}}} \right)}}^{2}}-{{{\left( {{{a}^{2}}-{{b}^{2}}} \right)}}^{2}}}}{{\left( {{{a}^{2}}-{{b}^{2}}} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)}}} \right]\div \left[ {\frac{{{{{\left( {a+b} \right)}}^{2}}-{{{\left( {a-b} \right)}}^{2}}}}{{\left( {a-b} \right)\left( {a+b} \right)}}} \right]\times \left( {\frac{{{{a}^{2}}+{{b}^{2}}}}{{ab}}} \right)\\\left[ {\because {{{\left( {x+y} \right)}}^{2}}-{{{\left( {x-y} \right)}}^{2}}=4xy} \right]\\=\frac{{4{{a}^{2}}{{b}^{2}}}}{{\left( {{{a}^{2}}-{{b}^{2}}} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)}}\div \frac{{4ab}}{{\left( {a-b} \right)\left( {a+b} \right)}}\times \frac{{\left( {{{a}^{2}}+{{b}^{2}}} \right)}}{{ab}}\\=\frac{{4{{a}^{2}}{{b}^{2}}}}{{\left( {{{a}^{2}}-{{b}^{2}}} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)}}\times \frac{{\left( {a-b} \right)\left( {a+b} \right)}}{{4ab}}\times \frac{{\left( {{{a}^{2}}+{{b}^{2}}} \right)}}{{ab}}\\=\frac{{4{{a}^{2}}{{b}^{2}}\left( {{{a}^{2}}+{{b}^{2}}} \right)\left( {{{a}^{2}}-{{b}^{2}}} \right)}}{{4{{a}^{2}}{{b}^{2}}\left( {{{a}^{2}}-{{b}^{2}}} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)}}\\=1\end{array} \)

(xii) \( \frac{{b+c}}{{bc}}\left( {b+c-a} \right)+\frac{{c+a}}{{ca}}\left( {c+a-b} \right)+\frac{{a+b}}{{ab}}\left( {a+b-c} \right) \)

সমাধানঃ

\(\begin{array}{l}\frac{{b+c}}{{bc}}\left( {b+c-a} \right)+\frac{{c+a}}{{ca}}\left( {c+a-b} \right)+\frac{{a+b}}{{ab}}\left( {a+b-c} \right)\\=\frac{{a\left( {b+c} \right)\left( {b+c-a} \right)}}{{abc}}+\frac{{b\left( {c+a} \right)\left( {c+a-b} \right)}}{{abc}}+\frac{{c\left( {a+b} \right)\left( {a+b-c} \right)}}{{abc}}\\=\frac{{a\left( {{{b}^{2}}+bc-ab+bc+{{c}^{2}}-ac} \right)}}{{abc}}+\frac{{b\left( {{{c}^{2}}+ac-bc+ac+{{a}^{2}}-ab} \right)}}{{abc}}+\frac{{c\left( {{{a}^{2}}+ab-ac+ab+{{b}^{2}}-bc} \right)}}{{abc}}\\=\frac{{a{{b}^{2}}+abc-{{a}^{2}}b+abc+a{{c}^{2}}-{{a}^{2}}c+b{{c}^{2}}+abc-{{b}^{2}}c+abc+{{a}^{2}}b-a{{b}^{2}}+{{a}^{2}}c+abc-a{{c}^{2}}+abc+{{b}^{2}}c-b{{c}^{2}}}}{{abc}}\\=\frac{{6abc}}{{abc}}\\=6\end{array}\)

(xiii) \( \frac{{{{y}^{2}}+yz+{{z}^{2}}}}{{\left( {x-y} \right)\left( {x-z} \right)}}+\frac{{{{z}^{2}}+zx+{{x}^{2}}}}{{\left( {y-z} \right)\left( {y-x} \right)}}+\frac{{{{x}^{2}}+xy+{{y}^{2}}}}{{\left( {z-x} \right)\left( {z-y} \right)}} \)

সমাধানঃ

\( \begin{array}{l}\frac{{{{y}^{2}}+yz+{{z}^{2}}}}{{\left( {x-y} \right)\left( {x-z} \right)}}+\frac{{{{z}^{2}}+zx+{{x}^{2}}}}{{\left( {y-z} \right)\left( {y-x} \right)}}+\frac{{{{x}^{2}}+xy+{{y}^{2}}}}{{\left( {z-x} \right)\left( {z-y} \right)}}\\=-\frac{{\left( {{{y}^{2}}+yz+{{z}^{2}}} \right)}}{{\left( {x-y} \right)\left( {z-x} \right)}}-\frac{{\left( {{{z}^{2}}+zx+{{x}^{2}}} \right)}}{{\left( {y-z} \right)\left( {x-y} \right)}}-\frac{{\left( {{{x}^{2}}+xy+{{y}^{2}}} \right)}}{{\left( {z-x} \right)\left( {y-z} \right)}}\\=-\left[ {\frac{{\left( {{{y}^{2}}+yz+{{z}^{2}}} \right)}}{{\left( {x-y} \right)\left( {z-x} \right)}}+\frac{{\left( {{{z}^{2}}+zx+{{x}^{2}}} \right)}}{{\left( {y-z} \right)\left( {x-y} \right)}}+\frac{{\left( {{{x}^{2}}+xy+{{y}^{2}}} \right)}}{{\left( {z-x} \right)\left( {y-z} \right)}}} \right]\\=-\left[ {\frac{{\left( {y-z} \right)\left( {{{y}^{2}}+yz+{{z}^{2}}} \right)}}{{\left( {x-y} \right)\left( {y-z} \right)\left( {z-x} \right)}}+\frac{{\left( {z-x} \right)\left( {{{z}^{2}}+zx+{{x}^{2}}} \right)}}{{\left( {y-z} \right)\left( {x-y} \right)\left( {z-x} \right)}}+\frac{{\left( {x-y} \right)\left( {{{x}^{2}}+xy+{{y}^{2}}} \right)}}{{\left( {z-x} \right)\left( {y-z} \right)\left( {x-y} \right)}}} \right]\\\left\{ {\because \left( {a-b} \right)\left( {{{a}^{2}}+ab+{{b}^{2}}} \right)={{a}^{3}}-{{b}^{3}}} \right\}\\=-\left[ {\frac{{{{y}^{3}}-{{z}^{3}}}}{{\left( {x-y} \right)\left( {y-z} \right)\left( {z-x} \right)}}+\frac{{{{z}^{3}}-{{x}^{3}}}}{{\left( {y-z} \right)\left( {x-y} \right)\left( {z-x} \right)}}+\frac{{{{x}^{3}}-{{y}^{3}}}}{{\left( {z-x} \right)\left( {y-z} \right)\left( {x-y} \right)}}} \right]\\=-\left[ {\frac{{{{y}^{3}}-{{z}^{3}}+{{z}^{3}}-{{x}^{3}}+{{x}^{3}}-{{y}^{3}}}}{{\left( {x-y} \right)\left( {y-z} \right)\left( {z-x} \right)}}} \right]\\=-\frac{0}{{\left( {x-y} \right)\left( {y-z} \right)\left( {z-x} \right)}}\\=0\end{array} \)