\(2,\\ a,=-3x^3y^3z^4\\ b,=\dfrac{1}{4}xy^2\cdot\dfrac{1}{4}x^4y^4\cdot\left(-\dfrac{4}{5}yz^2\right)=-\dfrac{1}{20}x^5y^7z^2\\ c,=-\dfrac{15}{14}x^6y^{11}z^{10}\\ 3,\\ a,=9\left(-1\right)\left(-\dfrac{1}{27}\right)=\dfrac{1}{3}\\ b,=-\dfrac{1}{5}\left(-8\right)=\dfrac{8}{5}\\ c,=\dfrac{4}{9}a\cdot36\left(-1\right)=-16a\)

1/6x(2y3)2.(-9.x5y)

= 1/6.x.4y6.(-9).x5.y

= [1/6.4.(-9)].(x.x5).(y6.y)

= -6.x6.y7

Đơn thức có bậc 6 + 7 = 13.

4xy2. [(-3)/4.(x2 y)3]

= 4xy2.[(-3)/4.x6.y3]

= [4.(-3)/4].(x.x6).(y2.y3)

= -3.x7.y5.

Đơn thức có bậc bằng 7 + 5 = 12.

\(\frac{\frac{2}{3}x^5y^3z}{\frac{4}{9}x^2y\left(\frac{1}{z^3}\right)}=\frac{\frac{2}{3}x^3.y^2}{\frac{2}{3}.\frac{1}{z^2}}=\frac{x^3.y^2}{z^2}\)

\(\Rightarrow\)  đơn thức bậc 3 

hệ số \(=1\)

a)\(4xy^2\) và \(\dfrac{3}{4}\left(x^2y\right)^3\)

= \(4xy^2\) . \(\dfrac{3}{4}\left(x^2y\right)^3\)

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

= \(3x^6y^5\)

b)\(\dfrac{1}{6}x\left(2y^3\right)^2\) và \(-9x^5y\)

= \(\dfrac{1}{6}x\left(2y^3\right)^2\) . \(-9x^5y\)

= \(\dfrac{1}{6}x.2y^5\) . \(-9x^5y\)

= \(\dfrac{1}{3}xy^5\) . \(-9x^5y\)

= \(-3x^6y^6\)

Bổ sung nha

a) \(4xy^2.\dfrac{3}{4}\left(x^2y\right)^3=3x^6y^5\)

Bậc của đơn thức \(3x^6y^5=11\)

b)\(\dfrac{1}{6}x\left(2y^3\right)^2.\left(-9x^5y\right)\) = \(-3x^6y^6\)

Bậc của đơn thức \(-3x^6y^6=12\)

b) \(\dfrac{1}{6}x\left(2y^3\right)^2.\left(-9x^5y\right)\)

=\(\dfrac{1}{6}x\left(4y^5\right).\left(-9x^5y\right)\)

= \(\dfrac{2}{3}xy^5.\left(-9x^5y\right)\)

= \(-6x^6y^6\)

\(E=\left(1\frac{1}{2}xy^2\right).\left(1\frac{1}{3}x^2y^3\right).\left(1\frac{1}{4}x^3y^4\right).....\left(1\frac{1}{2014}x^{2013}y^{2014}\right)\)

\(E=\left(\frac{3}{2}xy^2\right).\left(\frac{4}{3}x^2y^3\right).\left(\frac{5}{4}x^3y^4\right).....\left(\frac{2015}{2014}x^{2013}y^{2014}\right)\)

\(E=\left(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{2015}{2014}\right).\left(x.x^2.x^3......x^{2013}\right).\left(y^2y^3.y^4......y^{2014}\right)\)

\(E=\left(\frac{3.4.5......2015}{2.3.4......2014}\right).\left(x^{1+2+3+....+2013}\right).\left(y^{2+3+4+....+2014}\right)\)

\(E=\frac{2015}{2}.x^{2027091}.y^{2029104}\)

Đến đây tự kết luận nhé(hệ số;phần biến;đơn thức)

\(\dfrac{1}{4}.\left(x^2y^3\right)^2.\left(-2xy\right)\\ =\dfrac{1}{4}.x^4y^6.\left(-2xy\right)\\ =\left[\dfrac{1}{4}.\left(-2\right)\right].\left(x^4.x\right)\left(y^6.y\right)\\ =-\dfrac{1}{2}x^5y^7\)

Hệ số : `-1/2`

Bậc : `12`

\(a,\dfrac{1}{4}.\left(x^2y^3\right)^2.\left(-2xy\right)\\ =\dfrac{1}{4}.\left(-2\right).x^4.y^6.x.y\\ =-\dfrac{1}{2}x^5y^7\)
hệ số \(:-\dfrac{1}{2}\)

Bậc của đơn thức : \(12\)