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,\(\left(x-5\right)^2-16=\left(x-5\right)^2-4^2=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)

a) 3x³ + 6x²y

= 3x².(x + 2y)

b) 2x³ - 6x²

= 2x².(x - 2)

c) 18x² - 20xy

= 2x.(9x - 10y)

d) xy + y² - x - y

= (xy + y²) - (x + y)

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

= (x + y)(y - 1)

e) (x²y² - 8)² - 1

= (x²y² - 8 - 1)(x²y² - 8 + 1)

= (x²y² - 9)(x²y² - 7)

= (xy - 3)(xy + 3)(x²y² - 7)

f) x² - 7x - 8

= x² - 8x + x - 8

= (x² - 8x) + (x - 8)

= x(x - 8) + (x - 8)

= (x - 8)(x + 1)

a: \(3x^3+6x^2y\)

\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)

b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)

c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)

d: \(xy+y^2-x-y\)

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

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

e: \(\left(x^2y^2-8\right)^2-1\)

\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)

\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)

f: \(x^2-7x-8\)

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

\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)

g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)

\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)

\(=2x\left(2x-y\right)\left(5x-3y\right)\)

h: \(x^2-2x+1-y^2\)

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

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

i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)

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

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

\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)

l: \(-2x^2+8xy-8y^2\)

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

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

m: \(3x^2+5x-3y^2-5y\)

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

\(=3\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+3y+5\right)\)

g) 10x²(2x - y) + 6xy(y - 2x)

= 10x²(2x - y) - 6xy(2x - y)

= 2x(2x - y)(5x - 3y)

h) x² - 2x + 1 - y²

= (x² - 2x + 1) - y²

= (x - 1)² - y²

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

i) 2x(x + 2) + x² (-x - 2)

= 2x(x + 2) - x²(x + 2)

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

k) -9 + 6x - x²

= -(x² - 6x + 9)

= -(x - 3)²

l) 8xy - 2x² - 8y²

= -2(x² - 4xy + 4y²)

= -2(x - 2y)²

m) 3x² + 5x - 3y² - 5y

= (3x² - 3y²) + (5x - 5y)

= 3(x² - y²) + 5(x - y)

= 3(x - y)(x + y) + 5(x - y)

= (x - y)[3(x + y) + 5]

= (x - y)(3x + 3y + 5)

\(\left(x-5\right)^2-16=\left(x-5\right)^2-4^2=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)

\(25-\left(3-x\right)^2=5^2-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

\(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)\(49\left(y-4\right)^2-9\left(y+2\right)^2=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left(7y-28-3y-6\right)\left(7y-28+3y-6\right)=\left(4y-34\right)\left(10y-22\right)\)\(=4.\left(2y-17\right)\left(5y-11\right)\)

e); f) Áp dụng hằng đẳng thức số 6,7 để làm