Question

Tìm x

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{7}{x+8}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+7}-\dfrac{1}{x+8}=\dfrac{7}{x+8}\)

\(\Leftrightarrow\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{7}{x+8}\)

\(\Leftrightarrow8\left(x+8\right)=7x\left(x+8\right)\)

\(\Leftrightarrow\left(x+8\right)\left(8-7x\right)=0\)

=>x=-8(loại) hoặc x=8/7(nhận)