সরল কর:

$$\begin{gathered} \frac{x-y}{xy}+\frac{y-z}{yz}+\frac{z-x}{zx} \\ \frac{z(x - y) + x(y-z) + y(z-x)}{xyz} \end{gathered}$$

[এখানে, হর xy, yz ও zx এর ল.সা.গু. xyz]

$\frac{xz-yzxy-xz+yz-xy}{xyz}= \frac{0}{xyz}$

$$\begin{gathered} \frac{x-y}{(x+y)(y+z)}+ \frac{y-z}{(y+z)(z+x)}+\frac{z-x}{(z+x)(x + y)} \\ =\frac{(x - y)(z + x) + (y-z) (x + 1) + (z-x) (y + z)}{(x+y)(y+z) ( z+x)} \end{gathered}$$

এখানে, হরগুলে র ল.সা.গু. (x+y) (y+z)(z + x)

$\frac{zx + x² - yz - xy + xy + y² zx-yz+ yz + z² - xy - zx}{(x+y)\left(yz\right) (z+x)}$

$\frac{x²+ y² + z² - xy - yz-z}{(x+y)(y + z)(z + x)}$

$$\begin{gathered} \frac{y}{(x - y)(y - z)) }+\frac{ x}{(z - x)(x - y)} + \frac{z}{(y - z)(z - x)} \\ = \frac{y(z - x) + x(y - z) + z(x - y)}{(x - y)(y - z)(z - x)} \\ = \frac{(yz - xy + xy - zx + xz - yz)}{(x - y)(y - z)(z - x)} =\frac{ 0}{(x - y)(y - z)(z - x)} = 0 \end{gathered}$$

$$\begin{gathered} =\frac{1}{x+3y}+\frac{1}{x-3y}-\frac{2x}{x^2-9y^2} \\ =\frac{x-3y+x+3y}{(x+3y)(x-3y)}-\frac{2x}{x^2-9y^2}=\frac{2x}{x^2-9y^2}-\frac{2x}{x^2-9y^2}=0 \end{gathered}$$

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

$$\begin{gathered} =\left\{\frac{(x+y+xy)}{(x+y)(x+y)}\right\}-\left\{\frac{4x-2y+4x+2y}{(2x+y)(2x-y)}\right\} \\ =\frac{\left(2x\right)}{(x^2-y^2)}-\frac{\left(8x\right)}{(2x{)}^2-(y{)}^2}=\frac{2x}{\left(x^2{y}^2\right)}-\frac{8x}{4x^2-y^2} \\ =\frac{2x(4x^2-y^2)-8x(x^2-y^2)}{(x^2-y^2)\left(4x^{2}y\right)} \\ = \frac{6xy}{(x^2-y^2)\left(4x^{2}y\right)} \end{gathered}$$

প্রদত্ত রাশি  $=\frac{1}{x-2}-\frac{x-2}{x^2+2x+4}+\frac{6x}{x^3+8}$

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

                     $=\frac{6x}{(x{)}^3-(2{)}^3}+\frac{6x}{x^3+8}=\frac{6x}{x^3-8)}+\frac{6x}{(x^3+8)}$

                         $$\begin{gathered} =6x\left(\frac{1}{x^3-8}+\frac{1}{x^3+8}\right)=6x\left\{\frac{x^3+8+x^3-8}{(x^3-8)(x^3+8)}\right\} \\ =6x\times \frac{(2x^3}{(x^3{)}^2-(8{)}^2)}=\frac{12x^4}{x^6-64} \end{gathered}$$

প্রদত্ত রাশি $=\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}+\frac{4}{x^4+1}$

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

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

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

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

                =    $\frac{(x - y)² + (y - z)² + (z-x)²}{(x-y) (y-z)(z-x)}$

                =  $\frac{x² - 2xy + y² + y² - 2yz + z^2 + z²-2zx + x^2}{(x - y) (y-z) (z-x)}$

            =$\frac{2x^2+2y^2+2z^2-2xy-2yz-2zx}{(x-y)(y-z)(z-x)}$

           $=\frac{2(x^2+y^2+z^2*xy-yz-zx}{(x-y)(y-z)(z-x)}$

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

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

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

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

$$\begin{gathered} \frac{1}{a^2+b^2-c^2+2ab}+\frac{1}{b^2+c^2-a^2+2bc}+\frac{1}{c^2+a^2-b^2+2ca} \\ =\frac{1}{a^2+2ab+b^2-c^2}+\frac{1}{b^2+2bc+c^2-a^2}+\frac{1}{c^2+2ca+a^2-b^2} \\ =\frac{1}{(a+b{)}^2-(c{)}^2}+\frac{1}{(b+c{)}^2-(a{)}^2}+\frac{1}{(c+a{)}^2-(b{)}^2} \\ = \frac{1}{ (a +b+c)(a+b-c)} + \frac{1}{(b + c + a)(b + c - a)} + \frac{1}{(c + a + b)(c + a - b)} \\ = \frac{1}{(a + b + c)(a + b - c)}+ \frac{1}{(a + b + c)(b + c - a)}+\frac{ 1}{(a + b + c)(c + a - b)} \\ =\frac{ (b+c-a)(c+a-b)+(a+b-c)(c+a (a + f + c\left)\right(a + b - c\left)\right(b + c - a)}{(a+b-c)(b+c-a)(c+a-b)} \\ =\frac{2ab+2bc+2ca-a^2-b^2-c^2}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)} \end{gathered}$$