Giải bài 1.32 trang 18

Đề bài

Thực hiện phép chia:

a) $\left(4x^4{y}^2-6x^3{y}^3-2x^2{y}^4\right):\left(-2x^2{y}^2\right)$;       

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

Lời giải chi tiết

a) Ta có:

$\left(4x^4{y}^2-6x^3{y}^3-2x^2{y}^4\right):\left(-2x^2{y}^2\right)$

$=4x^4{y}^2:(-2x^2{y}^2)-6x^3{y}^3:(-2x^2{y}^2)-2x^2{y}^4:(-2x^2{y}^2)$

$=-2x^2+3xy+y^2.$

b) Ta có

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

$=5x^4{y}^3:\frac{5}{6}x{y}^2+\frac{1}{2}{x}^3{y}^4:\frac{5}{6}x{y}^2-\frac{2}{3}{x}^2{y}^5:\frac{5}{6}x{y}^2-x{y}^6:\frac{5}{6}x{y}^2$

$=6x^{3}y+\frac{3}{5}{x}^2{y}^2-\frac{4}{5}x{y}^3-\frac{6}{5}{y}^4$.