a, \(M=\dfrac{1}{2}x^4y^4\)

b, hệ số : 1/2 ; biến x^4y^4 ; bậc 8 

\(a,M=\left(\dfrac{2}{3}xy^3\right)\left(\dfrac{3}{4}x^3y\right)=\left(\dfrac{2}{3}.\dfrac{3}{4}\right)\left(x.x^3\right)\left(y^3.y\right)=\dfrac{1}{2}x^4y^4\)

b, Hệ số:\(\dfrac{1}{2}\)

Biến:x⁴y⁴

Bậc:8

gái mlem mlem :))

a: F=9/25x^2y^4*20/27x^3y=4/15x^5y^5

Bậc: 10

b: y=-x/3 và x+y=2

=>x+y=2 và -1/3x-y=0

=>x=3 và y=-1

Khi x=3 và y=-1 thì F=4/15*(-3)^5=-324/5

\(a,A=\left(\dfrac{-3}{8}x^2y\right)\left(\dfrac{2}{3}xy^2z^2\right)\left(\dfrac{4}{5}x^3y\right)\\ =\left(\dfrac{-3}{8}.\dfrac{2}{3}.\dfrac{4}{5}\right)\left(x^2.x.x^3\right)\left(y.y^2.y\right).z^2\\ =\dfrac{-1}{5}x^6y^4z^2\)

b, Hệ số: \(-\dfrac{1}{5}\)

Biến: \(x^6y^4z^2\)

c, Bậc: 12

d,Thay x=-1, y=-2, z=3 vào A ta có:
\(A=\dfrac{-1}{5}x^6y^4z^2=\dfrac{-1}{5}.\left(-1\right)^2.\left(-2\right)^4.3^2=\dfrac{-1}{5}.1.16.9=\dfrac{-144}{5}\)

\(a,B=\left(-\dfrac{1}{2}xy^3\right)\left(2x^3y\right)^2\)

\(=\left(-\dfrac{1}{2}.4\right)\left(x.x^6\right)\left(y^3.y^2\right)\)

\(=-2x^7y^6\)

\(b,x=2,y=\dfrac{1}{2}\Rightarrow B=-2.2^7.\dfrac{1}{2}^6=-4\)

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

b, Thay \(x=2;y=\dfrac{1}{2}\) vào B

\(B=\left(-2\right).2^7.\left(\dfrac{1}{2}\right)^5=-8\)

\(=\left(\dfrac{y}{x\left(y-2x\right)}-\dfrac{2}{y\left(y+1\right)-2x\left(y+1\right)}\right)\cdot\left(1+y\right)\)

\(=\left(\dfrac{y}{x\left(y-2x\right)}-\dfrac{2}{\left(y+1\right)\left(y-2x\right)}\right)\cdot\left(y+1\right)\)

\(=\left(\dfrac{y\left(y+1\right)-2x}{x\left(y-2x\right)\left(y+1\right)}\right)\cdot\dfrac{y+1}{1}\)

\(=\dfrac{y^2+2y-2x}{x\left(y-2x\right)}\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

M=\(6x^3-3xy^2+3+4x^3+12xy^2+7\)

=\(\left(6x^3+4x^3\right)-\left(3xy^2-12xy^2\right)+3+7\)

=\(10x^3+9xy^2+10\)

Thay x=-2,y=1/2 vào M:

\(10\cdot\left(-2\right)^3+9\cdot-2\cdot\left(\dfrac{1}{2}\right)^2+10\)

=10*-8+-18*1/4+10

=-80+-4.5+10

=-74.5

\(M=3\left(2x^3-xy^2+1\right)-4x\left(-x^2-3y^2\right)+7\)

\(M=6x^3-3xy^2+3+4x^3+12xy^2+7\)

\(M=10x^3+9xy^2+10\)

\(M=10\cdot\left(-2\right)^3+9\cdot\left(-2\right)\cdot0,5^2+7\)

\(M=-80-4,5+7\)

\(M=-74,5\)