Question

Cho x≠0,2,-2,Rút gọn rồi tính GTBT
A= (\(\dfrac{4}{x-2}\)-\(\dfrac{3}{x+2}\)) :\(\dfrac{x+14}{x^2}\). (Tại x= -3 )

Answer 1

\(A=\left(\dfrac{4}{x-2}-\dfrac{3}{x+2}\right):\dfrac{x+14}{x^2}\\ =\left(\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x^2}{x+14}\\ =\dfrac{4x+8-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\ =\dfrac{x+14}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\=\dfrac{x^2}{x^2-4}\)

Tại `x=-3` Ta có :

\(\dfrac{x^2}{x^2-4}\\ =\dfrac{-3^2}{-3^2-4}\\ =\dfrac{9}{9-4}\\ =\dfrac{9}{5}\)