a)B=3x³ -2y³-6x²y²+xy

   B=(3x³-6x²y²)+(xy-2y³)

   B=3x²(x-2y²)+y(x-2y²)

    B=(x-2y²)(3x²+y)
tại x=\(\frac{2}{3}\)và y=\(\frac{1}{2}\)ta có B=(x-2y²)(3x²+y)=(\(\frac{2}{3}\)-2*^{\(\frac{1}{2}\)}^2 )(3*\(\frac{2}{3}\)^2+\(\frac{1}{2}\))=\(\frac{1}{6}\)*\(\frac{11}{6}\)=\(\frac{11}{36}\)

b)C= 2x+xy²-x²y-2y

   C=(2x-2y)+(xy²-x²y)

   C=2(x-y)-xy(x-y)

   C=(2-xy)(x-y)

tại x=\(-\frac{1}{2}\)và y=\(-\frac{1}{3}\)ta có C=(2-xy)(x-y)=(2-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)

1)\(n^2\left(n-1\right)\left(n+1\right)-\left(n^2+2\right)\left(n^2-2\right)=n^2\left(n^2-1\right)-\left(n^4-4\right)=n^4-n^2-n^4+4\)

\(=-n^2+4\)

2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)

\(=y^4-81-y^4+16=-65\)

3)\(\left(x-2y+3\right)\left(x+2y-3\right)-\left(x-2y\right)\left(x+2y\right)=\left(x+3\right)^2-4y^2-\left(x^2-4y^2\right)\)

\(=x^2+6x+9-4y^2-x^2+4y^2=6x+9\)

4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

5)\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)

6)\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)

Học tốt nha bạn !

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\\\left(\dfrac{y}{3}\right)^2=\left(\dfrac{8}{27}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\\dfrac{y}{3}=\dfrac{8}{27}\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\\dfrac{y}{3}=-\dfrac{8}{27}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{8}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{8}{9}\end{matrix}\right.\end{matrix}\right.\)

c: =>8x-1=5

=>8x=6

hay x=3/4

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

m: (x-y)(x^2-2xy+y^2)

=(x-y)*(x-y)^2

=(x-y)^3

=x^3-3x^2y+3xy^2-y^3

n: =-(x^3+x^2y-x-x^2y-xy^2+y)

=-x^3+x+xy^2-y

o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)

=-x^3-x^2y^2+x^2+2xy+2y^3-2y

p: (1/2x-1)(2x-3)

=1/2x*2x-1/2x*3-2x+3

=x^2-3/2x-2x+3

=x^2-7/2x+3

q: (x-1/2y)(x-1/2y)

=(x-1/2y)^2

=x^2-xy+1/4y^2

r: (x^2-2x+3)(1/2x-5)

=1/2x^3-5x^2-x^2+10x+3/2x-15

=1/2x^3-6x^2+11,5x-15

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

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

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

Tại x = -5, y = 2 ta có :

\(A=\left[\left(-5\right)^2-2\right]+\left(-5\right)\left(2^2-1\right)=8\)

+) \(B=3x^3-2y^3-6x^2y^2\)

\(B=3x^3-\left(2y^3+6x^2y^2\right)=3x^3-2y^2\left(y+3x^2\right)\)

Tại x = 2/3, y = 1/2 ta có :

\(B=3.\left(\dfrac{2}{3}\right)^3-2.\left(\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}+3.\dfrac{4}{9}\right)=\dfrac{55}{36}\)

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

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

Tại x= -1/2, y = -1/3 ta có :

\(C=\left(\dfrac{-1}{2}\right)\left[2+\left(\dfrac{-1}{3}\right)^2\right]-\left(-\dfrac{1}{3}\right)\left[\left(\dfrac{-1}{2}\right)^2+1\right]=\left(-\dfrac{19}{18}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{-23}{36}\)

phần A viết nhầm : sửa đề

A=x^2y-y+xy^2-x

a: A=2/3x^2y+4x^2y=14/3x^2y

=14/3*9*7=294

b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16

c: C=x^3y^3(2+10-20)=-8x^3y^3

=-8*1^3(-1)^3=8

d: D=xy^2(2018+16-2016)

=18xy^2

=18(-2)*1/9=-4