Question

Answer 1

a: \(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)

b: \(A\left(4-2x\right)=2\) nên \(A=\dfrac{2}{4-2x}=\dfrac{1}{2-x}=\dfrac{-1}{x-2}\)

=>x-4=-1

hay x=3(nhận)

c: Để A là số nguyên thì \(x-2-2⋮x-2\)

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

hay \(x\in\left\{3;1;4;0\right\}\)