Bài 1 trang 16 SGK Toán 8 Tập 1 Cánh diều - CD

a) Ta có: (–xy)(–2x²y + 3xy – 7x)

= (–xy).(–2x²y) + (–xy).3xy – (–xy) . 7x

= 2x³y² – 3x²y² + 7x²y.

b) Ta có: $\left(\frac{1}{6}{x}^2{y}^2\right)(-0,3x^{2}y-0,4xy+1)$

\(\begin{align} & =\frac{1}{6}{{x}^{2}}{{y}^{2}}.(-0,3{{x}^{2}}y)-\frac{1}{6}{{x}^{2}}{{y}^{2}}.0,4xy+\frac{1}{6}{{x}^{2}}{{y}^{2}}.1 \\ & =\frac{-5}{6}{{x}^{4}}{{y}^{3}}-\frac{1}{15}{{x}^{3}}{{y}^{3}}+\frac{1}{6}{{x}^{2}}{{y}^{2}} \\ \end{align}\)

c) Ta có: (x + y)(x² + 2xy + y²)

= x.x² + x.2xy + x.y² + y.x² + y.2xy + y.y²

= x³ + 2x²y + xy² + x²y + 2xy² + y³

= x³ + (2x²y + x²y) + (xy²+ 2xy²) + y³

= x³ + 3x²y + 3xy² + y³.

d) Ta có: (x – y)(x² – 2xy + y²)

= x.x² – x.2xy + x.y² – y.x²– y.(– 2xy) – y.y²

= x³ – 2x²y + xy² – x²y + 2xy² – y³

= x³ – (2x²y + x²y) + (xy² + 2xy²) – y³

= x³ – 3x²y + 3xy² – y³.

-- Mod Toán 8 HỌC247