a/ Đkxđ của A là : \(\left\{{}\begin{matrix}x\ne\pm1\\x\ne\frac{1}{2}\end{matrix}\right.\)
Ta có :
\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{x-5}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{x+1+2\left(x-1\right)-\left(x-5\right)}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{2x+4}{1-2x}\) (do \(x\ne\pm1;x\ne\frac{1}{2}\))
Vậy \(A=\frac{2x+4}{1-2x}\) với đkxđ \(x\ne\pm1;x\ne\frac{1}{2}\)
b/ Ta có : \(A>0\)
\(\Leftrightarrow\frac{2x+4}{1-2x}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+4>0\\1-2x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+4< 0\\1-2x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\x< \frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\x>\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-2< x< \frac{1}{2}\\x\in\varnothing\end{matrix}\right.\)
Vậy...
a) A = \(\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
A = \(\left(\frac{1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right):\frac{1-2x}{\left(x+1\right)\left(x-1\right)}\)
A = \(\left(\frac{\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{1-2x}\)
A = \(\left(\frac{x+1+2x-2+5-x}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{1-2x}\)
A = \(\frac{2x+4}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{1-2x}\)
A = \(\frac{\left(2x+4\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(1-2x\right)}\)
A = \(\frac{2x+4}{1-2x}\)
b) Để A > 0 suy ra:
\(\frac{2x+4}{1-2x}>0\)
TH1:Cả tử cả mẫu dương
⇒2x + 4 > 0
⇔2x > -4
⇔x > -2
Và 1 - 2x > 0
⇔ -2x > -1
⇔ x < \(\frac{1}{2}\)
Nên ở TH1 thì nghiệm là -2 < x < \(\frac{1}{2}\)
TH2: Cả tử cả mẫu âm
⇒2x + 4 < 0
⇔2x < -4
⇔x < -2
Và 1- 2x < 0
⇔-2x < -1
⇔x > \(\frac{1}{2}\)
Nên ở TH2 thì không có nghiệm
Vậy để A > 0 thì x có nghiệm là -2 < x < \(\frac{1}{2}\)
Thêm ĐKXĐ : x ≠ -1; x ≠ 1; x ≠ \(\frac{1}{2}\)