a) \(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne\pm1\end{matrix}\right.\)
\(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x+1}{x\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}.\frac{x\left(x-1\right)}{x+1}\)
\(\Leftrightarrow A=\frac{x^2}{x-1}\)
b) Ta có :
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow x^2-x+\frac{1}{4}\ge0\)
\(\Leftrightarrow x^2-x+1>0\)
\(\Leftrightarrow x^2>x-1\)
\(\Leftrightarrow\frac{x^2}{x-1}>1\) \(\forall x\)
Vậy A > 1 với \(\forall x\) (\(x\ne0;x\ne\pm1\))
a) đkxđ
\(\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
Rgọn
A=\(\frac{x^2+1}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)
=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x^2+2-x^2}{x\left(x-1\right)}\)
=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}.\frac{x\left(x-1\right)}{x^2+1}\)
Vẫn tính dc nhưng kết quả hơi xấu..bạn xem lại coi có sai chỗ nào k nha
