Question

cho biểu thức A=^{\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}+\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)}
a) Tìm điều kiện xác định và rút gọn A
b) Tìm x để A>0

Answer 1

\(a,A=\dfrac{x+1+2-2x+5-x}{\left(1-x\right)\left(x+1\right)}\cdot\dfrac{\left(1-x\right)\left(x+1\right)}{2x-1}\left(x\ne1;x\ne-1;x\ne\dfrac{1}{2}\right)\\ A=\dfrac{8-2x}{2x-1}\\ b,A>0\Leftrightarrow\dfrac{8-2x}{2x-1}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-2x>0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}8-2x< 0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x< 4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 4\)