\(x^4-4x^3+9x^2-10x+6\)

\(=x^4-2x^3+3x^2-2x^3+4x^2-6x+2x^2-4x+6\)

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

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

a) \(=\left(a+2c\right)^2-16=\left(a+2c-4\right)\left(a+2c+4\right)\)

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

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

a, a² + 4ac + 4c² - 16 = (a + 2c)² - 4² = (a + 2c -4).(a + 2c +4)

b, 12y - 9x² + 36 - 3x²y = (12y + 36) - (3x²y + 9x²) = 12.(y+ 3) - 3x².(y + 3) =(y + 3).(12 - 3x²)

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

2, \(5x\left(3y+4x-6\right)\)

3, \(3x\left(2-y\right)\)

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

5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)

6, \(2xy\left(x+2y-5x^2y\right)\)

7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

11, \(\left(x+y\right)\left(x-1\right)\)

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

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

= (y - 2x)(4x + y)

e) (x + 5)² - (2x - 1)²

= (x + 5 - 2x + 1)(x + 5 + 2x - 1)

= (6 - x)(3x + 4)

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

(x - 10)(x + 2) = 0

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)

\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

a, \(x^4+2x^2+1-x^2\)

= \(\left(x^2+1\right)^2-x^2\)

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

b, \(x^4+x^2+1\)

= \(x^4+2x^2+1-x^2\)

= .. ( như phần a )

c, \(y^4+64\)

= \(\left(y^2+8\right)\left(y^2-8\right)\)

d, \(4xy+3z-12y-xz\)

\(=4y\left(x-3\right)-z\left(x-3\right)\)

\(=\left(x-3\right)\left(4y-z\right)\)

e, \(x^2-4xy+4y^2-z^2+6z-9\)

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

g, \(x^2-4xy+5x+4y^2-10y\)

\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)

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

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

h, \(x^2-7x+6\)

\(=x^2-6x-x+6\)

\(=x\left(x-6\right)-\left(x-6\right)\)

\(=\left(x-6\right)\left(x-1\right)\)

i, \(x^3+5x^2+6x+2\)

\(=x^3+x^2+4x^2+4x+2x+2\)

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

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

phần b là 6^4 nhé

nhầm câu c là 6^4

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

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

=4