Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

\(=x^2-3x+6x-12\)

\(=x^2+3x-12\)

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

\(=-2x^2y^2-2xy+3xy+3\)

\(=-2x^2y^2+xy+3\)

d: \(x\left(xy-1\right)\left(xy+1\right)\)

\(=x\left(x^2y^2-1\right)\)

\(=x^3y^2-x\)

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

Bài 3: 

Ta có: \(A=\left(x+2\right)\left(3x-1\right)-x\left(3x+3\right)-2x+7\)

\(=3x^2-x+6x-2-3x^2-9x-2x+7\)

=5

a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

a.

M=-3x²y+5

P=1/3x²y+xy²-xy+1/2xy²-5xy-1/3x²y

=(1/3x²y-1/3x²y)+(xy²-1/2xy²)-(5xy+xy)

=1/2xy²-6xy

Thay biểu thức P tại x=0,5 và y=1

=1/2.0,5.1²-6.0,5.1

=-11/4

Ta có P=1/3x²y+xy²-xy+1/2xy²-5xy-1/3x²y

P=(1/3x²y-1/3x²y)+(xy²+1/2xy²)+(-xy-5xy)

P=3/2xy²-6xy

Thay x=0,5 và y=1 ta có P=3/2.0,5.1²-6.0,5.1

P=0,75-3=-2.25          ( nha!)

P=1/3x²y+xy²-xy+1/2xy²-5xy-1/3x²y 

=(1/3x²y-1/3x²y)+(xy²+1/2xy²)-(xy-5xy)

=x²y+3/2xy²-6xy

=3/2xy²-6xy

Thay biểu thức P tại x=0,5 và y=1

3/2.1/5.1²-6.1/5.1

=-9/4

Vậy biểu thức P=-9/4 

\(Q=x^2+2xy+\left(-3x^3+3x^3\right)+\left(2y^3-y^3\right)=x^2+2xy+y^3\)

\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)

\(P=\dfrac{1}{3}x^2y+xy^2-xy+\dfrac{1}{2}xy^2-5xy-\dfrac{1}{3}x^2y=\dfrac{3}{2}xy^2-6xy\)

Thay x = 2 ; y = 1 ta được 

\(\dfrac{3}{2}.2.1-6.2.1=3-12=-9\)

A=3xyz - 3x^2 + 5xy - 1 B=5x^2 + 2xy^2 - 5 xy - y Tính 2A-B
`2A-B=6xyz-6x^2+10xy-2-(5x^2+2xy^2-5xy-y)`
`=6xyz+5xyz-6x^2-5x^2+10xy+5xy+y-2`
`=11xyz-11x^2+15xy+y-2`

tính A-B=(3xyz - 3x^2 + 5xy - 1) + (5x^2 + 2xy^2 - 5 xy - y)

             =3xyz - 3x^2 + 5xy - 1 + 5x^2 + 2xy^2 - 5 xy - y

             =3xyz+( 3x^2+ 5x^2 ) + (5xy- 5 xy)+2xy^2-1-y

             =3xyz+8x^2+2xy^2-1-y