Question

\(C=\dfrac{X^3}{X^2-4}-\dfrac{X}{X-2}-\dfrac{2}{2+X}\)

a) Rút gọn C

b) Tìm x để C = 0

c) Tìm giá trị nguyên của x để C nhận giá trị dương

Answer 1

\(\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

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

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

\(=x-1\)

b) \(C=0\Rightarrow x-1=0\Rightarrow x=1\)

c) C luôn dương => \(C\ge0\Rightarrow x-1\ge0\Rightarrow x\ge1\)