সমাধান কর: $\frac{1}{x+1}+\frac{1}{x+4}=\frac{1}{x+2}+\frac{1}{x+3}$

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

বা, $\frac{1}{x+1}-\frac{1}{x+2}=\frac{1}{x+3}-\frac{1}{x+4}$

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

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

বা, $\frac{1}{x^2+2x+x+2}=\frac{1}{x^2+4x+3x+12}$

বা, $\frac{1}{x^2+3x+2}=\frac{1}{x^2+7x+12}$

বা, $x^2+7x+12=x^2+3x+2$

বা, $x^2+7x-x^2-3x=+2-12$

বা, $4x=-10$

বা, $x=-\frac{10}{4}$

$\therefore x=-\frac52$

সুতরাং নির্ণেয় সমাধান $x=-\frac52$ [Answer]