Question

Rút gọn : A= \(\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9}\)

Tìm giá trị A khi \(\left|x-2\right|=1\)

Answer 1

A = \(\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9}\)

= \(\dfrac{5}{x+3}+\dfrac{2}{x-3}-\dfrac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

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

= \(\dfrac{-3x}{x+3}\)

TH1: x - 2 = 1

<=> x = 3

Thay x = 3 vào A, ta có:

A = \(\dfrac{-9}{6}\)

= \(\dfrac{-3}{2}\)

TH2: - x + 2 = 1

<=> x = 1

Thay x = 1 vào A, ta có:

A = \(\dfrac{-3}{4}\)