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

c,x^4-5x^2+4=x^4-4x^2-x^2+4=(x^2-4)(x^2-1)=(x-1)(x+1)(x-2)(x+2)

e,x^4-3x^3+x^2+3x-2=x^4-x^3-2x^3+2x^2-x^2+x+2x-2=(x-1)(x^3-2x^2-x+2)

Đến đây lấy máy tính bấm Mode*3+1+>+3 rồi tìm nghiệm

Các câu khác cũng máy tính đi

. Ai đó giúp tôi đi mà ._.

bài khó quá bạn ạ

1

a, 2x²+4x+2-2y² = 2(x²+2x+1-y²)= 2[(x+1)²-y² ] = 2(x-y+1)(x+y+1)

b, 2x - 2y - x² + 2xy - y²= 2(x -y) - (x² - 2xy + y²) = 2(x-y)-(x-y)²=(x-y)(2-x+y)

c, x²-y²-2y-1=x²-(y²+2y+1)=x²-(y+1)²=(x-y-1)(x+y+1)

d, x²-4x-2xy-4y+y²= x²-2xy+y²-4x-4y=(x-y)

2.

a, x²-3x+2=x²-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)

b, x²+5x+6=x²+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)

c, x²+6x-6=

a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

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

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

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

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

d: =2x^3+x^2+2x^2+x+2x+1

=(2x+1)(x^2+x+1)

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

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

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

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

b) \(2x^3-x^2+5x+3\)

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

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

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

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

c) \(2x^3-5x^2+5x+1\)

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

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

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

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

d) \(2x^3+3x^2+3x+1\)

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

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

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

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

e) \(3x^3-2x^2+5x+2\)

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

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

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

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

e) 27 - 3x²

= 3(9 - x²)

= 3(3 - x)(3 + x)

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

=>\(x\left(x+2\right)-3\left(x+2\right)=0\)

=>(x+2)(x-3)=0

=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

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

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

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

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

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) 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 = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu 

a) x² +5x -6 = x² -x +x + 5x -6

= x² -x +6x -6

= x( x-1) + 6(x-1) = (x-1)(x+6)

b) 5x² +5xy -x-y = 5x(x+y) -(x+y)

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

e) \(2x^2+3x-5\)

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

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

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

f) \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15+x-3\)

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

\(=\left(-5x+1\right)\left(x-3\right)\)

c)7x-\(6x^2\)-2

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

=(3x - \(6x^2\)) - (2 - 4x)

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

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

A = 6x⁴ - 5x³ + 4x² + 2x - 1

   = 6x⁴ + 3x³ - 8x³ - 4x² + 8x² + 4x - 2x - 1

   = 3x³. ( 2x + 1 ) - 4x² ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )

   = ( 2x + 1 ) ( 3x³ - 4x² + 4x - 1 )

    = ( 2x + 1 ) ( 3x³ - x² - 3x² + x + 3x - 1 )

     = ( 2x + 1 ) [ x² ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]

     = ( 2x + 1 ) ( 3x - 1 ) ( x² - x + 1 )

B = 4x⁴ + 4x³ + 5x² + 8x - 6

    = 4x⁴ - 2x³ + 6x³ - 3x² + 8x² - 4x + 12x - 6

     = 2x³ ( 2x - 1 ) + 3x² ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )

     = ( 2x - 1 ) ( 2x³ + 3x² + 4x + 6 )

     = ( 2x - 1 ) [ x² ( 2x + 3 ) + 2 ( 2x + 3 ) ]

      = ( 2x - 1 ) ( 2x + 3 ) ( x² + 2 )

C = x⁴ + x³ - 5x² + x - 6

   = x⁴ - 2x³ + 3x³ - 6x² + x² - 2x + 3x - 6 

   = x³ ( x - 2 ) + 3x² ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )

   = ( x - 2 ) ( x³ + 3x² + x + 3 )

    = ( x - 2 ) [ x² ( x + 3 ) + ( x + 3 ) ]

    = ( x - 2 ) ( x + 3 ) ( x² + 1 )