Question

[Lãng Tử Buồn]

Cho biểu thức Q=\(\left\{\frac{x^2-1}{x-1}+\frac{x^3-1}{1-x^2}\right\}:\frac{2x^2-4x+2}{x^2-1}\)

a, Rút gọn Q

b, tìm x sao cho \(|Q|\) > Q

Answer 1

[Hoàng Tử Hà]

a/ ĐKXĐ: \(x\ne1\)

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

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

\(Q=\frac{x}{2\left(x-1\right)}\)

b/ Để |Q|>Q

\(\Leftrightarrow Q\) <0

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\2\left(x-1\right)< 0\end{matrix}\right.\left(tm\right)\\\left\{{}\begin{matrix}x< 0\\2\left(x-1\right)>0\end{matrix}\right.\left(l\right)\end{matrix}\right.\Leftrightarrow0< x< 1\)

Answer 2

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ne1\)

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

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

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

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

\(\left|Q\right|>Q\Leftrightarrow Q< 0\Leftrightarrow\frac{x}{2\left(x-1\right)}< 0\Leftrightarrow0< x< 1\)