Question

Cho biểu thức

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

a) tìm ĐkXĐ rồi rút gọn B

b) Tìm x để B = 0

c) tìm x để b < 0

d) tìm x để b > 0

e)tìm x để B ≥ 5

Answer 1

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

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

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

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

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

b: Để B=0 thì \(x^2-2x-2=0\)

hay \(x\in\left\{1+\sqrt{3};1-\sqrt{3}\right\}\)