Question

Answer 1

a: ĐKXĐ: x<>2; x<>-2

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

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)

\(=\frac{2x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x^2-4+10-x^2}=\frac{-6}{6\left(x-2\right)}=\frac{-1}{x-2}\)

b: \(\left|x\right|=\frac12\)

=>\(\left[\begin{array}{l}x=\frac12\left(nhận\right)\\ x=-\frac12\left(nhận\right)\end{array}\right.\)

Thay \(x=\frac12\) vào A, ta được:

\(A=\frac{-1}{\frac12-2}=-1:\frac{-3}{2}=\frac23\)

Thay \(x=-\frac12\) vào A, ta được:

\(A=\frac{-1}{-\frac12-2}=-1:\frac{-5}{2}=\frac25\)

c: A<0

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

=>x-2>0

=>x>2

d: Để A là số nguyên thì -1⋮x-2

=>x-2∈{1;-1}

=>x∈{3;1}