Question

\(C=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{4-x^2}\right)\)

a) Rút gọn C

b) x bằng mấy để C = 1?

Answer 1

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

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

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

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

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

b: Để C=1 thì \(x^2+4=2-x\)

\(\Leftrightarrow x^2+x+2=0\)

hay \(x\in\varnothing\)