a: \(A=\dfrac{x-2}{x+2}+\dfrac{6x-4}{x^2-4}\)
\(=\dfrac{x-2}{x+2}+\dfrac{6x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)^2+6x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)
b: \(P=A+B=\dfrac{x}{x-2}+\dfrac{x+1}{x-2}=\dfrac{2x+1}{x-2}\)
Để P nguyên thì \(2x+1⋮x-2\)
=>\(2x-4+5⋮x-2\)
=>\(5⋮x-2\)
=>\(x-2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{3;1;7;-3\right\}\)
a.
\(A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{6x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-4x+4+6x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x}{x-2}\)
b.
\(P=A+B=\dfrac{x}{x-2}+\dfrac{x+1}{x-2}=\dfrac{2x+1}{x-2}=\dfrac{2\left(x-2\right)+5}{x-2}=2+\dfrac{5}{x-2}\)
Để P nguyên \(\Rightarrow\dfrac{5}{x-2}\in Z\)
\(\Rightarrow x-2=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x=\left\{-3;1;3;7\right\}\)
