Question

2/x-2 + ( 1/x+2 ) + ( 6+5x / 4 -x^2 )

Rút gọn biểu thức ( với x khác 2 và -2 )

Answer 1

\(\dfrac{2}{x-2}+\dfrac{1}{x+2}+\dfrac{6+5x}{4-x^2}\)

\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{6+5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x+4+x-2-6-5x}{\left(x-2\right)\left(x+2\right)}\)

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

= \(=\dfrac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-2}{x-2}\)

Answer 2

\(\dfrac{2}{x-2}+\dfrac{1}{x+2}+\dfrac{6+5x}{4-x^2}\\ =\dfrac{2\left(x+2\right)}{x^2-4}+\dfrac{x-2}{x^2-4}-\dfrac{6+5x}{x^2-4}\\ =\dfrac{2x+4+x-2-6-5x}{x^2-4}\\ =\dfrac{-2x-4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{2}{x-2}\)