Question

[Anh Vy]

giúp mk bài 6 và bài 7 với

Answer 1

[GV Nguyễn Trần Thành Đạt]

Bài 6:

\(a,2x^2y^2\left(x^3y^2-x^2y^3-\dfrac{1}{2}y^5\right)=2x^2y^2.x^3y^2-2x^2y^2.x^2y^3-2.x^2y^2.\dfrac{1}{2}y^5\\ =2x^5y^4-2x^4y^5-x^2y^7\\ b,-\dfrac{1}{3}xy\left(3x^2y^2-6x^2+y^2\right)\\ =-\dfrac{1}{3}.3.x.x^2.y.y^2-\dfrac{1}{3}.\left(-6\right)x.x^2.y-\dfrac{1}{3}x.y.y^2=-x^3y^3+2x^3y-\dfrac{1}{3}xy^3\\ c,\left(-2xy^2+\dfrac{2}{3}y^2+4xy^2\right).\dfrac{3}{2}xy=-2.\dfrac{3}{2}x.x.y.y^2+\dfrac{2}{3}.\dfrac{3}{2}xy.y^2+4.\dfrac{3}{2}x.x.y^2.y\\ =-3x^2y^3+xy^3+6x^2y^3=3x^2y^3+xy^3\)

Answer 2

[GV Nguyễn Trần Thành Đạt]

Bài 6:

\(d,x^2+2xy-3\left(-xy\right)=x^2+2xy+3xy=x^2+5xy\\ e,\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}x.y^2-1\right)=\dfrac{1}{2}.2.x^2.x^3.y-\dfrac{1}{2}.\dfrac{2}{5}.x^2y.x.y^2-\dfrac{1}{2}x^2y\\ =x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\\ f,\left(-xy^2\right)^2.\left(x^2-2x+1\right)=\left(xy^2\right)^2.\left(x-1\right)^2=\left[xy^2.\left(x-1\right)\right]^2\\ =\left(x^2y^2-xy^2\right)^2\)