Ta có:  9 x 2 - 1 = ( 3 x ) 2 - 1 2 = ( 3 x - 1 ) ( 3 x + 1 )

Ta có:  9 x 2 - 1 = ( 3 x ) 2 - 1 2 = ( 3 x - 1 ) ( 3 x + 1 )

\(a,Sửa:x^2+4xy-9+4y^2=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ b,=\left(x-6y\right)^2-1=\left(x-6y-1\right)\left(x-6y+1\right)\\ c,=36-\left(x-5y\right)^2=\left(6-x+5y\right)\left(6+x-5y\right)\)

a) \(x^3+3x=x\left(x+3\right)\)

b) \(9x^2-6x=3x\left(3x-2\right)\)

c) \(5y^{10}+15y^6=5y^6\left(y^4+3\right)\)

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

a) \(=x^2+7x-12x-84-2x+14\)

\(=x^2-7x-70\)

b)\(=x^2-4x-2x+8\)

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

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

c) \(=9x\left(x+y\right)-\left(x+y\right)\)

\(=\left(9x-1\right)\left(x+y\right)\)

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

\(=\left(x-y+9\right)\left(x-y-9\right)\)

e)\(=x^2+8x+16-60+15x\)

\(=x^2+23x-44\)

a) Ta có: \(\left(x-12\right)\left(x+7\right)-2x-14\)

\(=\left(x-12\right)\left(x+7\right)-2\left(x+7\right)\)

\(=\left(x+7\right)\left(x-14\right)\)

b) Ta có: \(x^2-6x+8\)

\(=x^2-2x-4x+8\)

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

c) Ta có: \(9x^2+9xy-\left(x+y\right)\)

\(=9x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(9x-1\right)\)

d) Ta có: \(\left(x^2-2xy+y^2\right)-81\)

\(=\left(x-y\right)^2-81\)

\(=\left(x-y-9\right)\left(x-y+9\right)\)

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

\(a,36-4x^2+20xy-25y^2\\ =36-\left(4x^2-20xy+25y^2\right)\\ =6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]\\ =6^2-\left(2x-5y\right)^2\\ =\left[6-\left(2x-5y\right)\right]\left[6+\left(2x-5y\right)\right]\\ =\left(6-2x+5y\right).\left(6+2x-5y\right)\)

a/

\(=6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]=\)

\(6^2-\left(2x-5y\right)^2=\left[6-\left(2x-5y\right)\right].\left[6+\left(2x-5y\right)\right]\)

\(=12x^2-6x=6x\left(2x-1\right)\)

`9x^2-4=(3x)^2-2^2=(3x-2)(3x+2)`

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

ADHĐT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;

b ) -(y-x-3)*(y+x+3)

a) \(12x^3-9x^2+3x\)

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

b) \(x^2-y^2+6x+9\)

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

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

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