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

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

\(=\frac{3}{4}\)

Ta có: \(3x^3-2y^3-6x^2y^2+xy\)

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

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

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

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

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

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

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, 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

bạn đặt nhân tử chung là xong bài rồi

a: \(M=(x−y)(x2−xy)−x(x2+2y2)M=(x−y)(x2−xy)−x(x2+2y2)\)

\(=x3−x2y−x2y+xy2−x3−2xy2=x3−x2y−x2y+xy2−x3−2xy2\)

\(=−2x2y−xy2=−2x2y−xy2\)

Khi x=2 và y=-3 thì \(M=−2⋅22⋅(−3)−2⋅(−3)2M=−2⋅22⋅(−3)−2⋅(−3)2\)

\(=−8⋅(−3)−2⋅9=24−18=6=−8⋅(−3)−2⋅9=24−18=6\)

b: \(N=x(x2+xy+y2)−y(x2+xy+y2)N=x(x2+xy+y2)−y(x2+xy+y2)\)

\(=(x2+xy+y2)(x−y)=x3−y3=(x2+xy+y2)(x−y)=x3−y3\)

Khi x=4 và y=-1 thì \(N=43−(−1)3=64+1=65N=43−(−1)3=64+1=65\)

c: \(P=a2(a+b)−b(a2−b2)+2013P=a2(a+b)−b(a2−b2)+2013\)

\(=a2(a+b)−b(a−b)(a+b)+2013=a2(a+b)−b(a−b)(a+b)+2013\)

\(=(a+b)(a2−ba+b2)+2013=a3+b3+2013=(a+b)(a2−ba+b2)+2013=a3+b3+2013\)

Khi a=1 và b=-1 thì \(P=13+(−1)3+2013=2013P=13+(−1)3+2013=2013\)