Giải bài 51 trang 58 – SGK Toán lớp 8 tập 1
Làm các phép tính sau:
a) \(\left( \dfrac{x^2}{y^2} + \dfrac{y}{x} \right) : \left(\dfrac{x}{y^2} - \dfrac{1}{y} + \dfrac{1}{x} \right)\)
b) \(\left( \dfrac{1}{x^2 + 4x + 4} - \dfrac{1}{x^2 - 4x + 4} \right) : \left(\dfrac{1}{x + 2} + \dfrac{1}{x - 2} \right)\)
a) \( \left( \dfrac{x^2}{y^2} + \dfrac{y}{x} \right) : \left(\dfrac{x}{y^2} - \dfrac{1}{y} + \dfrac{1}{x} \right)\)
\(= \left( \dfrac{x^3}{xy^2} + \dfrac{y^3}{xy^2} \right) : \left(\dfrac{x^2}{xy^2} - \dfrac{xy}{xy^2} + \dfrac{y^2}{xy^2} \right)\)
\(= \dfrac{x^3 + y^3}{xy^2} : \dfrac{x^2 - xy + y^2}{xy^2}\)
\(= \dfrac{x^3 + y^3}{xy^2}. \dfrac{xy^2}{x^2 - xy + y^2}\)
\(= \dfrac{(x^3 + y^3).xy^2}{xy^2.(x^2 - xy + y^2)}\)
\(= \dfrac{(x + y)(x^2 - xy + y^2).xy^2}{xy^2.(x^2 - xy + y^2)}\)
\(= x + y\)
b) \(\left( \dfrac{1}{x^2 + 4x + 4} - \dfrac{1}{x^2 - 4x + 4} \right) : \left(\dfrac{1}{x + 2} + \dfrac{1}{x - 2} \right)\)
\(= \left[ \dfrac{1}{(x + 2)^2} - \dfrac{1}{(x - 2)^2} \right] : \left(\dfrac{1}{x + 2} + \dfrac{1}{x - 2} \right)\)
\(= \left[ \dfrac{(x - 2)^2}{(x + 2)^2(x - 2)^2} - \dfrac{(x + 2)^2}{(x + 2)^2(x - 2)^2} \right] : \left[\dfrac{x - 2}{(x + 2)(x - 2)} + \dfrac{x + 2}{(x + 2)(x - 2)} \right]\)
\(= \dfrac{(x - 2)^2 - (x + 2)^2}{(x + 2)^2(x - 2)^2} : \dfrac{x - 2 + x + 2}{(x + 2)(x - 2)}\)
\(= \dfrac{(x - 2 - x - 2)(x - 2 + x + 2)}{(x + 2)^2(x - 2)^2} : \dfrac{2x}{(x + 2)(x - 2)}\)
\(= \dfrac{-4.(2x)}{(x + 2)^2(x - 2)^2}. \dfrac{(x + 2)(x - 2)}{2x}\)
\(= \dfrac{-4}{(x + 2)(x - 2)}\)
\(= \dfrac{-4}{x^2 - 4}\)
Lưu ý: \( \dfrac{A}{B}:\dfrac{C}{D} = \dfrac{A}{B}.\dfrac{D}{C}\)