Bài 3 trang 17 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

a) Ta có: 3x(2xy – 5x²y)

= 3x.2xy – 3x.5x²y

= (3.2).(x.x).y – (3.5).(x.x²).y

= 6x²y – 15x³y.

b) Ta có: 2x²y(xy – 4xy² + 7y)

= 2x²y.xy – 2x²y.4xy² + 2x²y.7y

= 2.(x².x).(y.y) – (2.4).(x².x).(y.y²) + (2.7).x².(y.y)

= 2x³y² – 8x³y³ + 14x²y².

c) $\left(\right)$Ta có: $\left(-\frac{2}{3}x{y}^2+6y{z}^2\right).\left(-\frac{1}{2}xy\right)$

$=-\frac{2}{3}x{y}^2.\left(-\frac{1}{2}xy\right)+6y{z}^2.\left(-\frac{1}{2}xy\right)$

$=\left[\left(-\frac{2}{3}\right).\left(-\frac{1}{2}\right)\right].\left(x.x\right).\left(y^2.y\right)+\left[6.\left(-\frac{1}{2}\right)\right].x.\left(y.y\right).z^2$

$=\frac{1}{3}{x}^2{y}^3-3x{y}^2{z}^2$.

-- Mod Toán 8 HỌC247