Question

[xube hoc ngu :33]

tim tich cua hai don thuc sau: 

2/3*x^2*y*z va -3*x*y^2

Answer 1

[Nguyễn Lê Phước Thịnh]

Ta có: \(\dfrac{2}{3}x^2yz\cdot\left(-3xy^2\right)\)

\(=\left(-\dfrac{2}{3}\cdot3\right)\cdot\left(x^2\cdot x\right)\cdot\left(y\cdot y^2\right)\cdot z\)

\(=-2x^3y^3z\)

Answer 2

[Yeutoanhoc]

`(2/3x^2.y.z).(-3x.y^2)`

`=-3.(2/3).x^2.y.z.x.y^2`

`=-1.x^{2+1}.y^{2+1}.z`

`=-x^3y^3z`