Đáp án+Giải thích các bước giải:
`ĐKXĐ:x\ne\{-1;-2;-3;-4;-5;-6;-7;-8\}`
Ta có:
`1/((x+1)(x+2))`
`=(x+2-x-1)/((x+1)(x+2))`
`=(x+2)/((x+1)(x+2))-(x+1)/((x+1)(x+2))`
`=1/(x+1)-1/(x+2)`
Tương tự ta có:
`1/((x+2)(x+3))=1/(x+2)-1/(x+3)`
`1/((x+3)(x+4))=1/(x+3)-1/(x+4)`
`1/((x+7)(x+8))=1/(x+7)-1/(x+8)`
`\to 1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))+...+1/((x+7)(x+8))=1/14`
`⇔1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+...+1/(x+7)-1/(x+8)=1/14`
`⇔1/(x+1)+(-1/(x+2)+1/(x+2))+(-1/(x+3)+1/(x+3))+(-1/(x+4)+1/(x+4))+...+(-1/(x+7)+1/(x+7))-1/(x+8)=1/14`
`⇔1/(x+1)-1/(x+8)=1/14`
`⇔(x+8-x-1)/((x+1)(x+8))=1/14`
`⇔7/((x+1)(x+8))=1/14`
`⇔(x+1)(x+8)=98`
`⇔x^2+9x+8=98`
`⇔x^2+9x-90=0`
`⇔x^2-6x+15x-90=0`
`⇔x(x-6)+15(x-6)=0`
`⇔(x-6)(x+15)=0`
\(⇔\left[ \begin{array}{l}x-6=0\\x+15=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=6(tm)\\x=-15(tm)\end{array} \right.\)
Vậy tập nghiệm của phương trình là: `S=\{6;-15\}`
