Question

Cho biểu thức: \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}-\dfrac{2-x^2}{x-x^2}\right)\). Tìm x để A<0

Answer 1

Đk: \(x\ne0,x\ne1\)

Ta có: \(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x^2-1+x+2-x^2}=\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)\(=\dfrac{x^2}{x-1}\)

Để A<0 \(\Leftrightarrow\dfrac{x^2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)   (vì \(x^2>0\))

Mà \(\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x\ne0\end{matrix}\right.\)