a. (2x - 5)²

b. x(x² - 1) = x(x - 1)(x + 1)

c. (x - 3)(x² + 3x + 9)

d. 5x(x - y) + (x - y) = (x - y)(5x + 1)

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

Ta có : 5x(x - 2y) + 2(2y - x)²

= 5x(x - 2y) + 2(x - 2y)² (vì (2y - x)² = (x - 2y)² )

= (x - 2y)[5x + 2(x - 2y)]

= (x - 2y)(5x + 2x - 4y)

= (x - 2y)(7x - 4y)

b) 7x(y - 4)² - (4 - y)³ 

= 7x(y - 4)² - (4 - y)²(4 - y)

= 7x(y - 4)² - (y - 4)²(4 - y)

= (y - 4)²(7x - 4 + y)

c) (4x - 8)(x² + 6) - (4x - 8)(x + 7) + 9(8 - 4x)

= (4x - 8)(x² + 6) - (4x - 8)(x + 7) - 9(4x - 8)

= (4x - 8)(x² + 6 - x - 7 - 9)

= 2(x - 4)(x² - x - 10)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

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

1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)

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

2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)

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

4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

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

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

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

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

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

\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)

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

1) 4x² - 7x - 2 = 4x² - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )

2) 4x² + 5x - 6 = 4x² - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )

3) 5x² - 18x - 8 = 5x² - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )

4) xy( x + y ) - yz( y + z ) + xz( x - z )

= x²y + xy² - y²z - yz² + xz( x - z )

= ( x²y - yz² ) + ( xy² - y²z ) + xz( x - z )

= y( x² - z² ) + y²( x - z ) + xz( x - z )

= y( x - z )( x + z ) + y²( x - z ) + xz( x - z )

= ( x - z )[ y( x + z ) + y² + xz ]

= ( x - z )( xy + yz + y² + xz )

= ( x - z )[ ( xy + y² ) + ( xz + yz ) ]

= ( x - z )[ y( x + y ) + z( x + y ) ]

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

5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )

\(4x^2-7x-2=\left(4x^2-8x\right)+\left(x-2\right)=4x\left(x-2\right)+\left(x-2\right)=\left(4x-1\right)\left(x-2\right)\)

\(=4x^2+8x-3x-6=4x\left(x+2\right)-3\left(x+2\right)=\left(4x-3\right)\left(x+2\right)\)

\(=5x^2-18x-8=5x^2-20x+2x-8=5x\left(x-4\right)+2\left(x-4\right)=\left(5x+2\right)\left(x-4\right)\)

\(5=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

a) = (x - 1)(5x - 3)

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

= (x + y)(x - 5)

c) = x(x - y) - y(x - y)

= (x - y)^2

d) = 2(x² + 2x + 1 - y²)

= 2[(x + 1)² - y²]

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