Question

Tìm x : \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{4}\)

Answer 1

Ezzz

ĐKXĐ: \(x\ne0;x\ne-2;x\ne\pm1\)

\(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}=\dfrac{3}{4}\)

<=> \(\dfrac{1}{x-1}-\dfrac{1}{x+2}=\dfrac{3}{4}\)

<=> \(\dfrac{x+2-x+1}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{3}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{12}{4\left(x-1\right)\left(x+2\right)}=\dfrac{3\left(x-1\right)\left(x+2\right)}{4\left(x-1\right)\left(x+2\right)}\)

<=> 12=3x²+3x-6

<=>3x²+3x-6-12=0

<=> 3x²+3x-18=0

<=> 3(x-2)(x+3)=0

<=> \(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) (thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là S={2;-3}