Question

Tính:

\(\dfrac{x}{x-2}+\dfrac{x-1}{x+2}-\dfrac{x-10}{4-x^2}\)

Answer 1

\(\dfrac{x}{x-2}+\dfrac{x-1}{x+2}-\dfrac{x-10}{4-x^2}\\ =\dfrac{x}{x-2}+\dfrac{x-1}{x+2}+\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x+x^2-2x-x+2+x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+x^2+2x-2x-x+x+2-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ =2\)

Answer 2

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x+x^2-3x+2+x-10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{x^2-4}=2\)