Bài 13:
a: ĐKXĐ: \(x\notin\left\{1;-1;-3\right\}\)
\(A=\left(\dfrac{7-2x}{x-1}+\dfrac{2x}{x+1}-\dfrac{1}{x^2-1}\right):\dfrac{3x+9}{x^2-1}\)
\(=\left(\dfrac{-2x+7}{x-1}+\dfrac{2x}{x+1}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{3\left(x+3\right)}\)
\(=\dfrac{\left(-2x+7\right)\left(x+1\right)+2x\left(x-1\right)-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{3\left(x+3\right)}\)
\(=\dfrac{-2x^2-2x+7x+7+2x^2-2x-1}{3\left(x+3\right)}\)
\(=\dfrac{3x+6}{3\left(x+3\right)}=\dfrac{x+2}{x+3}\)
b: |x-2|=1
=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Khi x=3 thì \(A=\dfrac{3+2}{3+3}=\dfrac{5}{6}\)
c: Để A nguyên thì \(x+2⋮x+3\)
=>\(x+3-1⋮x+3\)
=>\(-1⋮x+3\)
=>\(x+3\in\left\{1;-1\right\}\)
=>\(x\in\left\{-2;-4\right\}\)
Bài 12:
a: \(A=\left(\dfrac{1}{2-2x}+\dfrac{3}{2x+2}-\dfrac{2x^2}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)
\(=\left(\dfrac{-1}{2\left(x-1\right)}+\dfrac{3}{2\left(x+1\right)}-\dfrac{2x^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{-\left(x+1\right)+3\left(x-1\right)-4x^2}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{-x-1+3x-3-4x^2}{2}\cdot\dfrac{1}{-2x+1}\)
\(=\dfrac{-4x^2+2x-4}{2\left(-2x+1\right)}=\dfrac{-2x^2+x-2}{-2x+1}=\dfrac{2x^2-x+2}{2x-1}\)
b: Để A nguyên thì \(2x^2-x+2⋮2x-1\)
=>\(2⋮2x-1\)
=>\(2x-1\in\left\{1;-1;2;-2\right\}\)
=>\(2x\in\left\{2;0;3;-1\right\}\)
=>\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2}\right\}\)
Kết hợp ĐKXĐ và x nguyên, ta được:
\(x=0\)
