a) x²-xz-9y²+3yz

=(x²-9y²)-(xz-3yz)

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

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

b)x³-x²-5x+125

=x³-6x²+25x+5x²-30x+125

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

c.x³+2x²-6x-27

=x³+5x²+9x-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

d. 12x³+4x²-27x-9

=12x³+4x²-27x-9

=4x²(3x+1)-9(3x+1)

=(4x²-9)(3x+1)

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

e.x⁴-25x²+20x-4

=x⁴+5x³-2x²-5x²-25x+10+2x²+10x-4

=x²(x²+5x-2)-5(x²+5x-2)+2(x²+5x-2)

=(x²-5x+2)(x²+5x-2)

f.x²(x²-6)-x²+9

=x⁴+x³-3x²-x³-x²+3x-3x²-3x+9

=x²(x²+x-3)-x(x²+x-3)-3(x²+x-3)

=(x²-x-3)(x²+x-3)

mà bạn ơi sao câu b bạn tách ra nv ?

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

x2 nghĩa là x^2 hả bạn?

Giúp mik vs

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

a/

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

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

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) Ta có: \(1+6x-6x^2-x^3\)

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

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

b:Ta có: \(x^3-4x^2+8x-8\)

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

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

c: Ta có: \(x^3+2x^2+2x+1\)

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

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

d: Ta có: \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

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

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

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

\(=a^2\cdot\left(a+1\right)\left(a^3-a+2\right)\)

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

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

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

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

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

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

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

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³