Question

Giải phương trình: \(\dfrac{1}{x+1}\)+\(\dfrac{1}{x-2}\)≥\(\dfrac{1}{x-1}\)+\(\dfrac{1}{x}\)

Answer 1

ĐK: \(x\ne-1;0;1;2\)

\(\dfrac{1}{x-2}-\dfrac{1}{x-1}-\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)\ge0\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{1}{x\left(x+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{x^2+x-\left(x^2-3x+2\right)}{\left(x-2\right)\left(x-1\right)x\left(x+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{2\left(2x-1\right)}{\left(x-2\right)\left(x-1\right)x\left(x+1\right)}\ge0\)

Lập bảng xét dấu ta được nghiệm của BPT:

\(x\in\left(-1;0\right)\cup\text{[}\dfrac{1}{2};1\text{)}\cup\left(2;+\infty\right)\)