Giải bài 51 trang 58 - Sách giáo khoa Toán 8 tập 1

Đề bài

 Làm các phép tính sau:

a) \((\dfrac{x^2}{y^2}+\dfrac{y}{x}):(\dfrac{x}{y^2}-\dfrac{1}{y}+\dfrac{1}{x})\)

b) \((\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}):(\dfrac{1}{x+2}+\dfrac{1}{x-2})\)

 

Hướng dẫn giải

a) \((\dfrac{x^2}{y^2}+\dfrac{y}{x}):(\dfrac{x}{y^2}-\dfrac{1}{y}+\dfrac{1}{x})\)

= \(\dfrac{x^3+y^3}{xy^2}:\dfrac{x^2-xy+y^2}{xy^2}\) = \(\dfrac{x^3+y^3}{xy^2}.\dfrac{xy^2}{x^2-xy+y^2}\)

= \(\dfrac{(x+y)(x^2-xy+y^2)}{xy^2}\)\(.\dfrac{xy^2}{x^2-xy+y^2}\) = \(x+y\)

b) \((\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}):(\dfrac{1}{x+2}+\dfrac{1}{x-2})\)

= \([\dfrac{1}{(x+2)^2}-\dfrac{1}{(x-2)^2}]\)\(:(\dfrac{1}{x+2}+\dfrac{1}{x-2})\)

= \((\dfrac{1}{x+2}-\dfrac{1}{x-2})(\dfrac{1}{x+2}+\dfrac{1}{x-2}):(\dfrac{1}{x+2}+\dfrac{1}{x-2})\)

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