Bài tập 6.29 trang 12 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) $\frac{x^2-y^2}{6x^{2}y}:\frac{x+y}{3xy}=\frac{\left(x-y\right)\left(x+y\right)}{6x^{2}y}.\frac{3xy}{x+y}$

$=\frac{\left(x-y\right)\left(x+y\right).3xy}{6x^{2}y.\left(x+y\right)}=\frac{x-y}{2x}$.

b)$\mathrm{ 16}{x}^2{y}^2:\left(-\frac{18x^2{y}^5}{5}\right)=16x^2{y}^2.\frac{5}{\left(-18x^2{y}^5\right)}=\frac{16x^2{y}^2.5}{-18x^2{y}^5}=\frac{-40}{9y^3}$

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

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

$=\frac{\left(x-y\right)\left(x+y\right).3xy}{6x^{2}y.\left(x+y\right)}=\frac{x-y}{2x}$

-- Mod Toán 8 HỌC247