SOS
Bài 11:
a)ĐKXĐ : x khác 2,-2,0.
Ta có \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right)\times\dfrac{x^2+2x}{3x+3}=\left(\dfrac{x}{x^2-4}+\dfrac{5x+6}{x^2-4}\right)\times\dfrac{\left(x+2\right)x}{3x+3}=\dfrac{6x+6}{\left(x-2\right)\left(x+2\right)}\times\dfrac{\left(x+2\right)x}{3x+3}=\dfrac{2x}{x-2}\)
Vậy \(A=\dfrac{2x}{x-2}\)
b)Vì |x| = 1 => x ∈ {-1,1}
Với x = 1, ta có \(A=-2\)
Với x = -1, ta có \(A=\dfrac{2}{3}\)
c)Ta có \(A=\dfrac{2x}{x-2}=\dfrac{2x-4+4}{x-2}=2+\dfrac{4}{x-2}\) có GT nguyên <=> \(\dfrac{4}{x-2}\) có GT nguyên <=> 4 ⋮ x - 2 <=> x - 2 ∈ {-4,-2,-1,1,2,4} <=> x ∈ {-2,0,1,3,4,6}
Đối chiếu ĐKXĐ, ta có x ∈ {1,3,4,6}
