Question

[xube hoc ngu :33]

thu gon don thuc

(-x^3*z*y)*(2/3*y*x^2)^2

giup mk voi 

Answer 1

[Aikatsu]

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

\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)

\(=-\dfrac{4}{9}x^7.y^3.z\)

Answer 2

[Yeutoanhoc]

`(-x^3zy)(2/3yx^2)^2`

`=-4/9x^3zy.y^2x^4`

`=-4/9x^{3+4}.y^{1+2}z`

`=-4/9x^7y^3z`