Question

Answer 1

a: ĐKXĐ: x∉{2;-2}

b: \(A=\frac{6}{x+2}-\frac{2}{x-2}+\frac{x-10}{4-x^2}\)

\(=\frac{6}{x+2}-\frac{2}{x-2}-\frac{x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{6\left(x-2\right)-2\left(x+2\right)-x+10}{\left(x-2\right)\left(x+2\right)}=\frac{6x-12-2x-4-x+10}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x-6}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x+2}\)
c: Để A nguyên thì 3⋮x+2

=>x+2∈{1;-1;3;-3}

=>x∈{-1;-3;1;-5}