Question

[Chi Đào]

Thực hiện phép tính:

Answer 1

[Nguyễn Hoàng Minh]

\(a,\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^3}\\ =\dfrac{3\left(2x-1\right)}{x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-2x\right)\left(1+2x+4x^2\right)}\\ =\dfrac{-3\left(5x+1\right)}{x\left(1+2x+4x^2\right)}\)

\(b,\dfrac{x+3}{x^2-4}\cdot\dfrac{8-12x+6x^2-x^3}{9x+27}\\ =\dfrac{x+3}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(2-x\right)^3}{9\left(x+3\right)}\\ =\dfrac{-\left(2-x\right)^2}{9\left(x+2\right)}\)

\(c,\dfrac{x^2+5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\\ =\dfrac{x\left(x+2\right)\left(x+3\right)}{\left(x+4\right)\left(x-2\right)^2}\)

\(d,\dfrac{4a+6b}{a-1}:\dfrac{4a+12ab+9b^2}{1-a^3}\\ =\dfrac{2\left(2a+3b\right)}{a-1}\cdot\dfrac{\left(1-a\right)\left(a^2+a+1\right)}{\left(2a+3b\right)^2}\\ =\dfrac{-2\left(a^2+a+1\right)}{2a+3b}\)

\(e,\dfrac{x^2+2x-3}{x^2+3x-10}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+5\right)}\)