Question

tim tich cua hai don thuc 

-2*x^2*y^2*z va -1/3*x^2*y^3

Answer 1

\(\left(-2x^2y^2z\right).\left(-\dfrac{1}{3}x^2y^3\right)\)

\(=-2.x^2.y^2.z^2.\dfrac{-1}{3}.x^2.y^3\)

\(=\dfrac{2}{3}x^4y^5z^2\)

Answer 2

Ta có: \(-2x^2y^2z\cdot\dfrac{-1}{3}x^2y^3\)

\(=\left(-2\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(=\dfrac{2}{3}x^4y^5z\)

Answer 3

`(-2.x^2.y^2.z).(-1/3.x^2.y^3)`

`=-2/3.x^{2+2}.y^{2+3}.z`

`=-2/3x^4y^5x`