Ta phân tích :

`x^2+x=x(x+1)`

`x^2+3x+2=(x+1)(x+2)`

`x^2+5x+6=(x+2)(x+3)`

`x^2+7x+12=(x+3)(x+4)`

`x^2+9x+12=(x+4)(x+5)`

ĐKXĐ : `x\ne 0,x\ne -1,x\ne -2,x\ne -3,x\ne -4,x\ne -5`

`1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20)`

`=1/(x(x+1))+1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))+1/((x+4)(x+5))`

`=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)`

`=1/x-1/(x+5)`

`=(x+5-x)/(x(x+5))`

`=5/(x(x+5))`

Đáp án+Giải thích các bước giải:

$\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\\ =\dfrac{1}{x(x+1)}+\dfrac{1}{x^2+x+2x+2}+\dfrac{1}{x^2+2x+3x+6}+\dfrac{1}{x^2+3x+4x+12}+\dfrac{1}{x^2+4x+5x+20}\\ =\dfrac{1}{x(x+1)}+\dfrac{1}{x(x+1)+2(x+1)}+\dfrac{1}{x(x+2)+3(x+2)}+\dfrac{1}{x(x+3)+4(x+3)}+\dfrac{1}{x(x+4)+5(x+4)}\\ =\dfrac{1}{x(x+1)}+\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+\dfrac{1}{(x+3)(x+4)}+\dfrac{1}{(x+4)(x+5)}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\\ =\dfrac{1}{x}-\dfrac{1}{x+5}\\ =\dfrac{5}{x(x+5)}.$