\(=\left(x-y\right)\left(x+y\right)-10\left(x+y\right)=\)

\(=\left(x+y\right)\left(x-y-10\right)\)

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

= [( x + y) - 10)] . ( x - y)

  \(x^2\) - y² - 10\(x\) - 10y 

= (\(x^2\) - y²) - 10.(\(x\) + y)

= (\(x-y\))(\(x\) + y) - 10.(\(x\) + y)

= (\(x\) + y).(\(x\) - y - 10)

\(10x-25-x^2=-\left(x^2-10x+25\right)\)

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

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²

Đặt x²+10x+5=a.Ta có biểu thức là : a(a+8)+16=a²+8a+16=(a+4)²=(x²+10x+9)²

a) 6x² + 7xy + 2y²

= 6x² + 4xy + 3xy + 2y²

= (6x² + 4xy) + (3xy + 2y²)

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

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

b) x² - y² + 10x - 6y + 16

= x² + 10x + 25 - y² - 6y - 9

= (x² + 10x + 25) - (y² + 6y + 9)

= (x + 5)² - (y + 3)²

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

= (x - y + 2)(x + y + 8)

c) 4x⁴ + y⁴

= 4x⁴ + 4x²y² + y⁴ - 4x²y²

= (2x² + y²)² - (2xy)²

= (2x² + y² - 2xy)(2x² + y² + 2xy)

\(x^2-y^2+10x-6y+16\)

\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

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

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

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

1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2, \(x^2-10x+25=\left(x-5\right)^2\) 

3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2) \(x^2-10x+25=\left(x-5\right)^2\)

3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)

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

1: \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2: \(x^2-10x+25=\left(x-5\right)^2\)

3: \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

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

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

\(x^2+8x+16=\left(x+4\right)^2\)

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

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