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

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

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

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

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

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

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

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

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

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

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

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

2𝑥(𝑥+2)−(𝑥+2)(𝑥−2)

2𝑥^2+4𝑥−(𝑥+2)(𝑥−2)

2𝑥^2+4𝑥−(𝑥(𝑥−2)+2(𝑥−2))

2𝑥^2+4𝑥−(𝑥^2−2𝑥+2(𝑥−2))

2𝑥^2+4𝑥−(𝑥^2−2𝑥+2𝑥−4)

2𝑥^2+4𝑥−(𝑥^2−4)

2𝑥^2+4𝑥−𝑥^2+4

2𝑥^2−𝑥^2+4𝑥+4

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

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