\(-3ab.\left(a^2-3b\right)\)

\(=-3a^3b+9ab^2\)

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

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

\(=x^3-4x^2y+5xy^2-2y^3\)

a) \(-3ab.\left(a^2-3b\right)=-3ab.a^2+3ab.3b=-3a^3b+9ab^2\)

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

\(=xx^2-2yx^2-2xyx+2xy2y+xy^2-2yy^2\)

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

\(=x^3-4yx^2+5xy^2-2y^3\)

mk chỉ có thể thu gọn đc thôi, mk ko tính đc đâu!

a, \(2x\left(x-2y\right)-2y\left(y-2x\right)=2x^2-4xy-2y^2+4xy=2x^2-2y^2\)

b, \(\left(x-2\right)\left(3x^2+4x-5\right)=3x^3+4x^2-5x-6x^2-8x+10=3x^3-2x^2-13x+10\)

c, \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=x^3-2x^2y+4xy^2+2x^2y-4xy^2+8y^3=x^3+8y^3\)

a.

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

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

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

= 2x^2 + 2y^2

b.

(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)

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

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

= -4xy

a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)

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

\(=5x^3-7x^2y+2xy^2+5x-2y\)

b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

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

c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)

\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)

Answer:

\(3x^2.\left(2x^3-x+5\right)\)

\(=3x^2.2x^3+3x^2.(-x)+3x^2.5\)

\(=6x^5-3x^3+15x^2\)

\((4xy+3y-5x).x^2y\)

\(=4xy.x^2y+3y.x^2y-5x.x^2y\)

\(=4x^3+3x^2y^2-5x^3y\)

`a)A=x(x+y)-x(y-x)`

`=x^2+xy-xy+x^2`

`=2x^2`

Thay `x=-3`

`=>A=2.9=18`

`b)B=4x(2x+y)+2y(2x+y)-y(y+2x)`

`=8x^2+4xy+4xy+2y^2-y^2-2xy`

`=8x^2+y^2+6xy`

Thay `x=1/2,y=-3/4`

`=>B=8*1/4+9/16-9/4`

`=2+9/16-9/4`

`=9/16-1/4=5/16`

a)\({-1\over 2}x^2×y^2 - x^2×y^2 +{2\over 3} x^2×y^2 \)

=\(({ -1\over 2}-1+{ 2\over 3})x^2×y^2\)

=\({-5 \over 6}x^2×y^2\)

b)\({1 \over 2}a^3×b^2 +{4 \over 3}3ab^2 × {1 \over 2}a^2\)

=\({1 \over 2}a^3×b^2 +({4 \over 3}× {1 \over 2})3b^2 (a×a^2) \)

=\({1 \over 2}a^3×b^2 +{2 \over 3}3a^3b^2\)

=\(({1 \over 2} +{2 \over 3}3)a^3b^2\)

=\({5 \over 2}a^3b^2\)

c)