যোগফল নির্ণয় কর

প্রদত্ত রাশি = $= \frac{a - b}{a}+ \frac{a + b}{b} = \frac{b(a - b) + a(a + b)}{ab}$ [এখানে, হর a, b এর ল.সা.গু. ab]

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

$\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}$

প্রদত্ত রাশি = $\frac{x-y}{x}+\frac{y-z}{y}+\frac{z-x}{z}$

=$$\begin{gathered} \frac{yz(x - y) + zx(y - z) + xy(z - x)}{xyz} \\ =\frac{xyz - yz + xyz+xz² + xyz - x²y}{xyz} \\ =\frac{3xyz-x²y-yz-z²x}{xyz} \end{gathered}$$

প্রদত্ত রাশি= $=\frac{x+y}{x-y}+\frac{x-y}{x+y}+\frac{{(x+y)}^2+{(x-y)}^2}{(x+y)(x-y)}$ এখানে, হর (x - y), (x + y) এর ল.সা.গু. (x + y)(x - y)]$=\frac{x^2+2xy+y^2+x^2-2xy+y^2}{(x+y)(x-y)}=\frac{2x^2+2y^2}{x^2-y^2}=\frac{2(x^2+y^2)}{(x^2-y^2)}$

$\frac{1}{x^2-3x+2}+\frac{1}{x^2-4x+3}+\frac{1}{x^2-5x+4}$

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

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

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

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

এখানে, হর এর হর (x - 2)(x - 1) ,(x-3)(x-1) (x-4)(x-1)

$=\frac{x^2-7x+12+x^2-6x+8+x^2-5x+6}{(x-1)(x-2)(x-3)(x-4)}$

$=\frac{(3x^2-18x+26)}{(x-1)(x-2)(x-3)(x-4))}$

প্রদত্ত রাশি = $=\frac{1}{(a^2-b^2)}+\frac{1}{(a^2+ab+b^2)}+\frac{1}{(a^2-ab+b^2)}$

$=\frac{1}{(a+b)(a-b)}+\frac{1}{a^2+ab+b^2}+\frac{1}{a^2-ab+b^2}$

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

এখানে, হর (a + b) (a - b), a² + ab + b² ও a² – ab + b²  এর ল.সা.গু. (a + b)(a - b)(a² + ab + b²)(a² - ab + b²) ]

$\frac{\{(a²+b²)-ab\left]\{\right(a²+b²)+ab)+(a²+ab-ab-b²)+(a²-ab\text{'}+a\text{'}b-b\text{'})}{(a+b) (a-b) (a²+ab+b²) (a-ab+b²)}$$\frac{(a²+b²)²=a²b²+(a²+ab-ab-b)+(a²-a{b}^3+a^{3}b-{b^4}^{)}}{(a²+b²) (a²-b²)}$

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

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

$$\begin{gathered} =\frac{1}{(x-2)}+\frac{1}{(x+2)}+\frac{4}{(x^2-4)} \\ =\frac{1}{(x-2)}+\frac{1}{(x+2)}+\frac{4}{(x+2)(x-2)} \\ =\frac{x+2+x-2+4}{(x-2)(x+2)} \\ =\frac{2x+4}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}=\frac{2}{x-2} \end{gathered}$$

$\frac{1}{x²-1}+\frac{1}{x²-1}+\frac{4}{x^8-1}+\frac{1}{(x²)² - \left(1\right)²}+\frac{4}{ x^8 - 1}$

$=\frac{1}{(x^2-1)}+\frac{1}{(x^2+1)(x^2-1)}+\frac{4}{x^8-1}$

$=\frac{ x^2+1+1}{(x^2+1)(x^2-1)} +\frac{4}{x^8-1}=\frac{x^2+2}{(x^4-1)}+\frac{4}{(x^4{)}^2-(1{)}^2}$

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

$=\frac{(x^6+x^2+2x^4+2+4)}{(x^4{)}^2-(1{)}^2}=\frac{(x^6+2x^4+x^2+6)}{(x^8-1)}$