কষে দেখি 15 | 15. বীজগাণিতিক সংখ্যামালার সরলীকরণ | WBBSE Board Class 8 Math Solution

15. বীজগাণিতিক সংখ্যামালার সরলীকরণ | কষে দেখি 15 | Exercise 15 | Ganit Prabha Class VIII math solution | WBBSE Class 8 Math Solution in Bengali

 গণিত প্রভা VIII কষে দেখি 15 সমাধান 

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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 

\(=\frac{a}{c}+\frac{b}{c}\)  

\(=\frac{a+b}{c}=\) বামপক্ষ 

∴ সম্পর্কটি সত্য

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

সমাধানঃ

বামপক্ষ 

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

ডানপক্ষ 

\(=\frac{a}{x}+\frac{b}{y}\)  

\(=\frac{ay+bx}{xy}\neq\) বামপক্ষ 

∴ 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

∴ সম্পর্কটি সত্য

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

সমাধানঃ

বামপক্ষ 

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

\(=\frac{y+x}{xy}\)

\(\neq\frac{1}{x+y}\) 

সুতরাং, বামপক্ষ \(\neq\) ডানপক্ষ

∴ সম্পর্কটি মিথ্যা

2. নীচের বীজগাণিতিক ভগ্নাংশগুলি লঘিষ্ঠ আকারে প্রকাশ করি।

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

সমাধানঃ

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

\(=\frac{9a^3}{{11b}^{5-4}}\)

\(=\frac{9a^3}{11b}\)

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

সমাধানঃ

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

\(=\frac{6b^{5-2}c^2}{{7a}^{7-4}}\)

\(=\ \frac{6b^3c^2}{{7a}^3}\)

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

সমাধানঃ

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

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

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

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

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

সমাধানঃ

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

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

\(=\frac{a+1}{a-1}\times\frac{a\left(a+1\right)-2\left(a+1\right)}{a\left(a+1\right)}\) 

\(=\frac{1}{\left(a-2\right)}\times\frac{\left(a+1\right)\left(a-2\right)}{a}\)

\(=\frac{a+1}{a}\)

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

সমাধানঃ

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

\(=\frac{p^3+q^3}{p^2-q^2}\times\frac{p-q}{p+q}\)

\(=\frac{\left(p+q\right)\left(p^2+pq+q^2\right)}{\left(p+q\right)\left(p-q\right)}\times\frac{p-q}{p+q}\)

\(=\frac{(p^2-pq+q^2}{(p+q)}\)

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

সমাধানঃ

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

\(=\frac{x\left(x-3\right)+2\left(x-3\right)}{x(x+5)-1(x+5)}\times\frac{x\left(x+5\right)+1\left(x+5\right)}{x\left(x-3\right)-1\left(x-3\right)}\)

\(=\frac{\left(x-3\right)\left(x+2\right)}{(x+5)(x-1)}\times\frac{\left(x+5\right)\left(x+1\right)}{\left(x-1\right)(x-3)}\)

\(=\frac{x+2}{x-1}\times\frac{x+1}{x-1}\) 

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

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

 

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

সমাধানঃ

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

\(=\frac{(a^2-ab+b^2)}{a^2+ab}\times\frac{a^3-b^2}{a^3+b^3}\)

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

 \(=\frac{a^2-ab+b^2}{a(a+b)}\times\frac{a-b}{a^2-ab+b^2}\)

\(=\frac{(a-b)}{a(a+b)}\)\(=\frac{(a-b)}{(a^2-ab)}\)

3. নীচের বীজগাণিতিক ভগ্নাংশগুলি সরলতম আকারে প্রকাশ করি।

(i) \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

সমাধানঃ

\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

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

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

সমাধানঃ

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

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

\(=\frac{2a}{a}\)\(=2\)

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

সমাধানঃ

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

\(=\frac{{x(x}^2+a^2)+b\left(x-a\right)-ax.x^3}{abx}\)

\(=\frac{x^3+a^2x+bx-ab-ax^4}{abx}\)

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

সমাধানঃ

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

\(=\frac{2a^2b}{3b^2c}\times\frac{c^4}{3a^3}\times\frac{9a^2}{4bc^3}\)

\(=\frac{a}{2b^2}\)

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

সমাধানঃ

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

\(=\frac{1}{x^2-2x-x+2}+\frac{1}{x^2-2x-3x+6}\)

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

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

                                        \(+\frac{1}{x(x-3)-1(x-3)}\)

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

                                        \(+\frac{1}{(x-3)(x-1)}\)

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

\(=\frac{3x-6}{(x-2)(x-1)(x-3)}\)

\(=\frac{3(x-2)}{(x-2)(x-1)(x-3)}\)

\(=\frac{3}{(x-1)(x-3)}\) 

\(=\frac{3}{x^2-3x-x+3}\) 

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

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

সমাধানঃ

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

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

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

\(=\frac{2x\left(x^2+1\right)+2x(x^2-1)}{(x^2-1)(x^2+1)}+\frac{4x^3}{x^4+1}\)

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

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

\(=\frac{4x^3\left(x^4+1\right)+4x^3\left(x^4-1\right)}{(x^4-1)(x^4+1)}\)

\(=\frac{4x^7+4x^3+4x^7-4x^3}{(x^4-1)(x^4+1)}\)\(=\frac{8x^7}{x^8-1}\)

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

সমাধানঃ

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

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

\(=\frac{b(b-5)}{3b-4a}\times\frac{{(3b)}^2-{(4a)}^2}{b^2-5^2}\times\frac{a(b+5)}{b(3b+4a)}\)

\(=\frac{b\left(b-5\right)}{3b-4a}\times\frac{\left(3b+4a\right)\left(3b-4a\right)}{\left(b+5\right)\left(b-5\right)}\times\frac{a\left(b+5\right)}{b\left(3b+4a\right)}\)

\(=a\)

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

সমাধানঃ

\(\frac{b+c}{\left(a-b\right)(a-c)}+\frac{c+a}{\left(b-a\right)(b-c)}+\frac{a+b}{\left(c-a\right)\left(c+b\right)}\)

\(=\frac{b+c}{\left(a-b\right)(a-c)}-\frac{c+a}{\left(a-b\right)(b-c)}+\frac{a+b}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b+c\right)\left(b-c\right)-\left(a+c\right)\left(a-c\right)+(a+b)(a-b)}{(a-b)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{b^2-c^2-\left(a^2-c^2\right)+\left(a^2-b^2\right)}{(a-b)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{b^2-c^2-a^2+c^2+a^2-b^2}{(a-b)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{0}{(a-b)\left(a-c\right)\left(b-c\right)}\)\(=0\)

 

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

সমাধানঃ

  \(\frac{b+c-a}{\left(a-b\right)(a-c)}+\frac{c+a-b}{\left(b-c\right)(b-a)}+\frac{a+b-c}{\left(c-a\right)(c-b)}\)

\(=-\frac{b+c-a}{\left(a-b\right)(c-a)}-\frac{c+a-b}{\left(b-c\right)(a-b)}-\frac{a+b-c}{\left(c-a\right)(b-c)}\)

\(=-\left[\frac{b+c-a}{\left(a-b\right)(c-a)}+\frac{c+a-b}{\left(b-c\right)(a-b)}+\frac{a+b-c}{\left(c-a\right)(b-c)}\right]\)

\(=-\left[\frac{\left(b+c-a\right)\left(b-c\right)+\left(c+a-b\right)\left(c-a\right)+(a+b-c)(a-b)}{\left(a-b\right)(c-a)(b-c)}\right]\)

\(=-\left[\frac{0}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\right]\)

\(=0\)

(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

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

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

\(=\ \frac{\frac{a^2+ax-a^2}{x-a}+\frac{b^2+bx-b^2}{x-b}+\frac{c^2+cx-c^2}{x-b}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

\(=\ \frac{\frac{ax}{x-a}+\frac{bx}{x-b}+\frac{cx}{x-b}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

\(=\ \frac{x\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-b}\right)}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

\(=\ x\)