a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

Bạn viết rõ hơn nhé : 

\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)

= \(\frac{x^4-xy^3}{2xy+y^2}.\frac{2x+y}{x^3+x^2y+xy^2}\)

= \(\frac{x.\left(x-y\right).\left(x^2+xy+y^2\right).\left(2x+y\right)}{y.\left(2x+y\right).x.\left(x^2+xy+y^2\right)}\)

= \(\frac{x-y}{y}\)

Chúc bạn học tốt !!!

1, ( x + 3 )( x- 4 ) + ( x - 4 ) mũ 2

=x^2+4x+3x-12+x^2-8x+16

=2x^2-x+4

3, x( x -14 ) - 10(x - 1) mũ 2

=x^2-14x-10(x^2-2x+1)

=x^2-14x-10x^2-20x+10

=-9x^2-34x+10

a, \(\left(x+3\right)\left(x-4\right)-\left(x-4\right)^2=x^2-x-12-x^2+8x-16=7x-28\)

b, \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)=x^3-6x^2+12x-9+3x-3x^2+x^3\)

\(=2x^3-9x^2+15x-9\)

c, \(x\left(x-4\right)-10\left(x-1\right)^2=x^2-4x-10\left(x^2-2x+1\right)\)

\(=x^2-4x-10x^2+20x-10=-9x^2+16x-10\)

d, \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)=2x^2+4x-x^2+4=x^2+4x+4=\left(x+2\right)^2\)

a)(x+5)³-15x(x+10)

=x³+15x²+75x+125-15x²-150x

=x³+75x+125

b) (x-2)²-(x-5)²

=(x-2-x+5)(x-2+x-5)

=3.(2x-7)

=6x-21

c)(x+2)(x²-2x+4)-(x³+8)

=(x³+8)-(x³+8)

=0

#H

(x + 2)(x - 2) - (x - 2)(x + 5)

= (x - 2)(x + 2 - x - 5)

= (x - 2)-3

= -3x + 6

b) 2x(3x²y + 4x²y - 3)

= 2x(7x²y - 3)

= 14x³y - 6x

bạn giải hết đc ko ạ

Bài 9:

a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)

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

\(=3xy-y^2\)

\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)

b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Bài 7: 

a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)

b) \(93\cdot107=100^2-7^2=10000-49=9951\)

c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)

d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)

e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1=1\)

f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)

Bài 8:

a) Ta có: \(\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)

\(=\left(x^2+3x+1-3x+1\right)^2\)

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

\(=x^4+4x^2+4\)

b) Ta có: \(\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)

\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)

\(=-9x^2\)

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

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

\(=-4x^2\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)

\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)

\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)

\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)

\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)

a) (x-3)(x²+3x+9)-(x²-1)(x+27)

=(x³-27)-(x³+27x²-x-27)

=x³-27-x³-27x²+x+27

=-27x²+x

=x(-27x+1)

b) (3-x)³-(x+3)(x²-3x+9)

=27-27x+9x²-x³-x³-27

=-2x³+9x²-27x

=x(-2x+9x-27)

c) (x-2)(x²+2x+4)-x(x-3)(x+3)

=x³-8-x(x²-9)

=x³-8-x³+9x

=9x-8

#H