Bài tập 6.28 trang 12 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) $\frac{x^2-6x+9}{x^2-3x+9}.\frac{x^3+27}{3x-9}=\frac{{\left(x-3\right)}^2}{x^2-3x+9}.\frac{\left(x+3\right)\left(x^2-3x+9\right)}{3\left(x-3\right)}$

$=\frac{{\left(x-3\right)}^2\left(x+3\right)\left(x^2-3x+9\right)}{3\left(x-3\right)\left(x^2-3x+9\right)}=\frac{\left(x-3\right)\left(x+3\right)}{3}=\frac{x^2-9}{3}$.

b) $\frac{2x^2-20x+50}{3x+3}.\frac{x^2-1}{4{\left(x-5\right)}^3}$

$=\frac{2\left(x^2-10x+25\right)}{3\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{4{\left(x-5\right)}^3}$

$=\frac{2{\left(x-5\right)}^2}{3\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{4{\left(x-5\right)}^3}$

$=\frac{2{\left(x-5\right)}^2\left(x-1\right)\left(x+1\right)}{3\left(x+1\right).4{\left(x-5\right)}^3}=\frac{x-1}{6\left(x-5\right)}$

-- Mod Toán 8 HỌC247