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-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)

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

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

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

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

Ta có: \(3x^3-2y^3-6x^2y^2+xy\)

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

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

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

\(=\left(\dfrac{2}{3}-2\cdot\dfrac{1}{4}\right)\cdot\left(3\cdot\dfrac{4}{9}+\dfrac{1}{2}\right)\)

\(=\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\cdot\left(\dfrac{4}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{11}{6}=\dfrac{11}{36}\)

Bài 1:

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

Voqis x=-1;y=3 ta có:

\(A=\left(-1+3\right)\left(-1\cdot3-1\right)=2\cdot\left(-4\right)=-8\)

b) \(B=x^2y^2+xy+x^3+y^3=\left(x^2y^2+x^3\right)+\left(xy+y^3\right)\\ =x^2\left(y^2+x\right)+y\left(x+y^2\right)=\left(x+y^2\right)\left(x^2+y\right)\)

Với x=-1;y=3 ta có:

\(B=\left(-1+3^2\right)\left(-1^2+3\right)=8\cdot2=16\)

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

Với x=-1;y=3 ta có:

\(C=\left(-1-3\right)\left(2-\left(-1\right)\cdot3\right)=-4\cdot5=-20\)

d) phân tích tt

Ta có: H = x³ + x²y - xy² - y³ + x² - y² + 2x + 2y + 4 

= x²(x + y) - y²(x + y) + (x² - y²) + 2(x + y + 2)

= (x + y)(x² - y²) + (x² - y²) + 2(x + y + 1 + 1)

= (x + y + 1)(x² - y²) + 2(0 + 1)

= 0(x² - y²) + 2.1

= 2

Vậy H = 2

Chúc bn học tốt!

Help mik lẹ với ;-;

a) \(A=x^2y+y+xy^2-x\) (hẳn đề là vậy)

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

\(A=\left(-5\right).2\left(-5+2\right)+2+5\)

\(A=30+7=37\)

b) \(B=3x^3-2y^3-6x^2y^2+xy\)

\(B=3.\left(\frac{2}{3}\right)^3-2.\left(\frac{1}{2}\right)^3-6.\left(\frac{2}{3}\right)^2.\left(\frac{1}{2}\right)^2+\frac{2}{3}.\frac{1}{2}\)

\(B=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}\)

\(B=\frac{11}{36}\)

c) \(C=2x+xy^2-x^2y-2y\)

\(C=2.\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right).\left(-\frac{1}{3}\right)^2-\left(-\frac{1}{2}\right)^2.\left(-\frac{1}{3}\right)-2.\left(-\frac{1}{3}\right)\)

\(C=-1-\frac{1}{18}+\frac{1}{12}+\frac{2}{3}\)

\(C=-\frac{11}{36}\)

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