Question

thực hiện phép tính

\(\dfrac{12x^3}{y^4}.\dfrac{2y^5}{x^2}\)

\(\dfrac{5x-10}{4x-8}:\dfrac{2x+4}{4-2x}\)

\(\dfrac{3x+10}{x+3}.\dfrac{2x+4}{x+3}\)

Answer 1

Giải:

a) \(\dfrac{12x^3}{y^4}.\dfrac{2y^5}{x^2}\)

\(=\dfrac{12.2.x^3.y^5}{x^2.y^4}\)

\(=\dfrac{24.x^2.x.y^4.y}{x^2.y^4}\)

\(=24xy\)

Vậy ...

b) \(\dfrac{5x-10}{4x-8}:\dfrac{2x+4}{4-2x}\)

\(=\dfrac{5\left(x-2\right)}{4\left(x-2\right)}:\dfrac{2\left(x+2\right)}{2\left(2-x\right)}\)

\(=\dfrac{5}{4}:\dfrac{x+2}{2-x}\)

\(=\dfrac{5}{4}.\dfrac{2-x}{x+2}\)

\(=\dfrac{5\left(2-x\right)}{4\left(x+2\right)}\)

\(=\dfrac{10-5x}{4x+8}\)

Vậy ...

c) \(\dfrac{3x+10}{x+3}.\dfrac{2x+4}{x+3}\)

\(=\dfrac{3x+10}{x+3}.\dfrac{2\left(x+2\right)}{x+3}\)

\(=\dfrac{\left(3x+10\right).2.\left(x+2\right)}{\left(x+3\right)^2}\)

Vậy ...