Question

[33. Nguyễn Minh Ngọc]

Cho biểu thức: A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

a) Rút gọn biểu thức A
b) Tìm giá trị của x để A < 0

Answer 1

[l҉o҉n҉g҉ d҉z҉]

a) ĐKXĐ : x ≠ ±2

\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\div\frac{6}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=\frac{-1}{x-2}\)

b) Để A < 0 thì -1/x-2 < 0

=> x - 2 > 0 <=> x > 2

Vậy với x > 2 thì A < 0