Bài 3:
a: Thay x=-3 vào P, ta được:
\(P=\dfrac{-3+3}{-3-2}=\dfrac{0}{-5}=0\)
b: \(Q=\dfrac{x-1}{x+2}+\dfrac{5x-2}{x^2-4}\)
\(=\dfrac{x-1}{x+2}+\dfrac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-2\right)+5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-3x+2+5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)
c: Đặt A=Q/P
\(=\dfrac{x}{x-2}:\dfrac{x+3}{x-2}=\dfrac{x}{x+3}\)
Để A là số nguyên thì \(x⋮x+3\)
=>\(x+3-3⋮x+3\)
=>\(-3⋮x+3\)
=>\(x+3\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-2;-4;0;-6\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\varnothing\)
