Bài 3:
ĐKXĐ: \(x\notin\left\{1;-1;\dfrac{1}{2}\right\}\)
a: \(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(=\left(-\dfrac{1}{x-1}+\dfrac{2}{x+1}-\dfrac{x-5}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{2}{-2x+1}=\dfrac{-2}{2x-1}\)
b: Để A nguyên thì \(-2⋮2x-1\)
=>\(2x-1\in\left\{1;-1;2;-2\right\}\)
=>\(2x\in\left\{2;0;3;-1\right\}\)
=>\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2}\right\}\)
mà x nguyên và \(x\notin\left\{1;-1;\dfrac{1}{2}\right\}\)
nên x=0
c: |A|=A
=>\(\left|-\dfrac{2}{2x-1}\right|=\dfrac{-2}{2x-1}\)
=>\(-\dfrac{2}{2x-1}>=0\)
=>2x-1<0
=>2x<1
=>x<1/2
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x\ne-1\end{matrix}\right.\)
Bài 4:
ĐKXĐ: \(x\notin\left\{-1;0;2\right\}\)
a: \(Q=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}-\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right)\cdot\dfrac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\left(\dfrac{2}{x^2-x+1}-\dfrac{2}{x+1}\right)\cdot\dfrac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\dfrac{2x+2-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\dfrac{-2x^2+4x}{x+1}\cdot\dfrac{1}{x\left(x-2\right)}\)
\(=1+\dfrac{-2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\dfrac{2}{x+1}=\dfrac{x-1}{x+1}\)
b: \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Thay x=-1/2 vào Q, ta được:
\(Q=\dfrac{-\dfrac{1}{2}-1}{-\dfrac{1}{2}+1}=\dfrac{-3}{2}:\dfrac{1}{2}=-3\)
c: Để Q nguyên thì \(x-1⋮x+1\)
=>\(x+1-2⋮x+1\)
=>\(-2⋮x+1\)
=>\(x+1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{0;-2;1;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;1;-3\right\}\)
