dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

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

\(=\left(2x+y\right).3y\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

\(=\left(3x-2y\right)\left[3x+2y-1\right]\)

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

\(=x\left(x-y\right)^2-4x\)

\(=x\left[\left(x-y\right)^2-4\right]\)

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

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x²-y²+x+y+x-y)=2(x²-y²+2x)=2x²-2y²+4x

2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)

b.=(3x)²+2.3x.y+y²-(2z)²=(3x+y)²-(2z)²=(3x+y-2z)(3x+y+2z)

c.=x²-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)

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

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

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

\(=4x^2\)

hk tốt

^^

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

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

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

a) \(x^4-y^4\)

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

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

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

\(=0\cdot0\)

\(=0\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

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

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

\(=\left(x-5y\right)\left(5x-y\right)\)

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

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

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

\(=3x\left(x-2\right)\)

f) \(x^3+27\)

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

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

g) \(27x^3-0,001\)

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

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

a) (x - 1)(x + l)(x - 2)(x - 4).      b) (x - 2)( x 2  + 4).

c) 2y(3 x 2   +   y 2 ).                          d) 2(x + y + z) ( a   -   b ) 2 .

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

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

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

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

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

b. \(x^3-2x^2+4x-8\)

\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)

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

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

c. \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

\(=6x^2y+2y^3\)

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

d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)

\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)

\(=2\left(a-b\right)^2\left(x+y+z\right)\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu