Question

Rút gọn: C= \(\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}.\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{4-x^2}\right)\)

Answer 1

C=\(\dfrac{1}{x+2}-\dfrac{x\left(x-2\right)\left(x+2\right)}{x^2+4}.\left[\dfrac{1}{\left(x+2\right)^2}+\dfrac{1}{\left(x-2\right)\left(x+2\right)}\right]\)

C=\(\dfrac{1}{x+2}-\dfrac{x\left(x-2\right)\left(x+2\right)}{x^2+4}.\left[\dfrac{x-2}{\left(x-2\right)\left(x+2\right)^2}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)^2}\right]\)

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

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

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

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

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

C=\(\dfrac{2-x}{x^2+4}\)