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a) \(\frac{1}{3}\left(xy\right)^3.\left(-2x\right).\left(\frac{-3}{5}\right)y^5z\)
\(=\frac{1}{3}.\left(-2\right).\left(\frac{-3}{5}\right).x^3.x.y^3.y^5.z\)
\(=\frac{2}{5}.x^4y^8z\)
b) \(\left(\frac{-1}{3}\right)x^2yz.\frac{1}{7}\left(xy\right)^4.\frac{7}{9}xyz^3\)
\(=\left(\frac{-1}{3}\right).\frac{1}{7}.\frac{7}{9}.x^2.x^4.x.y.y^4.y.z.z^3\)
\(=\frac{-1}{27}.x^7y^6z^4\)
Giải:
a) \(\dfrac{1}{3}\left(xy\right)^3.\left(-2\right)x.\dfrac{-3}{5}y^5z\)
\(=\dfrac{1}{3}.\left(-2\right).\dfrac{-3}{5}x^3y^3xy^5z\)
\(=\dfrac{2}{5}x^4y^8z\)
Vậy ...
b) \(-\dfrac{1}{3}x^2yz.\dfrac{1}{7}\left(xy\right)^4.\dfrac{7}{9}xyz^3\)
\(=-\dfrac{1}{3}.\dfrac{1}{7}.\dfrac{7}{9}x^2yz.x^4y^4.xyz^3\)
\(=-\dfrac{1}{27}x^7y^6z^4\)
Vậy ...
a/ \(\frac{-1}{5}x^3y^2\frac{5}{4}xy^3\)
\(=\left(\frac{-1}{5}.\frac{5}{4}\right)\left(x^3.x\right)\left(y^2.y^3\right)\)
\(=\frac{-1}{4}x^4y^5\)
b/ \(-3xy^4\left(-\frac{1}{3}\right)x^2y\)
\(=\left(-3.\frac{-1}{3}\right)\left(x.x^2\right)\left(y^4y\right)\)
\(=1x^3y^5\)
a, 7\(xyz^2\) - 9\(xy\)z2 + \(\dfrac{1}{2}\)\(xyz^2\)
= \(xyz^2\).( 7 - 9 + \(\dfrac{1}{2}\))
=-\(\dfrac{3}{2}\) \(xyz^2\)
b, \(\dfrac{8}{3}\)\(xy\) - \(\dfrac{1}{4}\)\(xy\) + 25\(xy\)
= \(xy\).(\(\dfrac{8}{3}\) - \(\dfrac{1}{4}\) + 25)
=\(\dfrac{329}{12}\) \(xy\)