Question

Answer 1

Bài 8:

a: ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right):\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{x^2-4+10-x^2}\)

\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}\)

\(=\dfrac{2x-2-2x-4}{\left(x-2\right)}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)

b: |x|=1/2

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Khi x=1/2 thì \(A=\dfrac{-1}{\dfrac{1}{2}-2}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)

Khi x=-1/2 thì \(A=-\dfrac{1}{-\dfrac{1}{2}-2}=-1:\dfrac{-5}{2}=\dfrac{2}{5}\)

c: A=2

=>\(\dfrac{-1}{x-2}=2\)

=>\(x-2=-\dfrac{1}{2}\)

=>\(x=\dfrac{3}{2}\left(nhận\right)\)

d: A<0

=>\(-\dfrac{1}{x-2}< 0\)

=>x-2>0

=>x>2

e: Để A nguyên thì \(-1⋮x-2\)

=>\(x-2\in\left\{1;-1\right\}\)

=>\(x\in\left\{3;1\right\}\)