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=3.(2/3)^3-2.(1/2)^3-6.(2/3)^2.(1/2)^2+(2/3).(1/2)

      =8/9-1/4-2/3+1/3=8/9-1/4-1/3=11/36

b,  B=-1+(-1/18)+1/12+2/3=-11/36

a) A=xy(x+y) - (x+y) = (x+y) (xy-1) = (-5+2) (-5.2 -1) =-3 . -11 = 33
b) B= xy (y-x)+2(x-y) =xy (y-x) - 2(y-x) =(y-x) (xy -2)= (-1/3 - -1/2) ( -1/2 . -1/3 -- 2)= 1/6 . -11/6 =-11/ 36

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

Thay x = 1 và y = 1/2 vào biểu thức ta có

(1^2)*((1/2)^3) + 1*1/2

= 1/8 + 1/2

= 1/8 + 4/8

= 5/8

5/8 bn nhé

HT

(1^2)*((1/2)^3) + 1*1/2

= 1/8 + 4/8

= 5/8