Question

thu gon don thuc

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

giup mk voi 

Answer 1

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

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

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

Answer 2

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

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

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

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

Answer 3

\(\left(-x^3zy\right)\left(\dfrac{2}{3}yx^2\right)^2\)

\(\Leftrightarrow\left(-x^3yz\right)\left(\dfrac{2}{3}x^2y\right)\left(\dfrac{2}{3}x^2y\right)\)

\(\Leftrightarrow-\left[x^7y^3z\left(\dfrac{2}{3}\right)^2\right]\)

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