a) \(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

c) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

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

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

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

b) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

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

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

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

1.

= (x^3 + 125 ) -(x^2 +5x)

=(x +5) (x^2 -5x +25) -x(x+5)

=(x+5)(x^2 -5x +25 -x)

=(x+5)(x^2 -6x +25)

2.

= (x^3 -27) + (2x^2 -6x)

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

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

=(x-3) (x^2 +5x +9)

a) \(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

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

b) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

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

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

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

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

c) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)-\left(2x^2-6x\right)\)

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

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

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

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

e) \(4x^4+4x^3-x^2-x\)

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

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

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

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

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

\(=\left(x-1\right)\left[x^4\left(x+1\right)-9x^2\right]\)

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

Mình sửa: Bài 1
2)x²+3x-15

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)

\(x^2-7x+10\)

\(=x^2-2x-5x+10\)

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

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

học tốt

a ) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

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

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

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

b ) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

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

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

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

a) x³ - x² - 5x + 125

=(x³-6x²+25x)+(5x²-30x+125)

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

b) x³ + 2x² - 6x - 27

=x³+5x²+9-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

a) \(2x^3+6xy-x^2z-3yz\)

= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)

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

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

b)\(x^2-6xy+9y^2-49\)

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

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

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

Cứu với mn ơi , mai em có ktra ạ 

câu b nè : ta có  x³  - x² - 5x + 125 = x ^3 + 5* x^2 - 6x^2 - 30x + 25 x + 125 = x^2 (x+5 ) - 6x (x + 5) + 25 (x+5) 

= (x+ 5 ) (x^2 - 6x + 25)

nha bạn

câu a nè bạn :

có cl í tin người vcl !!!

a) x³ - x² - 5x + 125

=(x³-6x²+25x)+(5x²-30x+125)

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

b) x³ + 2x² - 6x - 27

=x³+5x²+9-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

c) 12x³ + 4x² - 27x - 9

=4x²(3x+1)-9(3x+1)

=(4x²-9)(3x+1)

=[(2x)²-3²](3x+1)

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

a) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

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

\(=\left(x+5\right)\left(x^2-5x+25-x\right)=\left(x+5\right)\left(x^2-6x+25\right)\)

b) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

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

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

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

c) \(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x-1\right)\left(4x^2-9\right)=\left(3x-1\right)\left(2x-3\right)\left(2x+3\right)\)

a) \(x^3-x^2-5x+125\)

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

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

\(=\left(x+5\right)\left(x^3-6x+25\right)\)