1. \(x^2+3x+2=x^2+2x+x+2=x\left(x+2\right)+\left(x+2\right)=\left(x+2\right)\left(x+1\right)\)
2.
\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(3x-1\right)\)3 và 4 bạn xem lại đề nha mik giải ko đc
5. \(x^2+2x-8=x^2+4x-2x-8=x\left(x+4\right)-2\left(x+4\right)=\left(x+4\right)\left(x-2\right)\)

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

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

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) \(27x^3-27x^2+18x-4\)

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

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

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

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

1)x²-3x+2

=x²-x-2x+2

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

=(x-1)(x-2)

2)2x²+5x-7

=2x²-2x+7x-7

=2x(x-1)+7(x-1)

=(x-1)(2x+7)

3)x²+7x-8

=x²-x+8x-8

=x(x-1)+8(x-1)

=(x-1)(x+8)

4)x²+8x+7

=x²+x+7x+7

=x(x+1)+7(x+1)

=(x+1)(x+7)

5)2x²-5x-7

=2x²+2x-7x-7

=2x(x+1)-7(x+1)

=(x+1)(2x-7)

xong rồi !!

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

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

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

b) 2x² - x - 6 = 2x² - 4x + 3x - 6 = 2x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 2x + 3 )

c) x² - 5x - 6 = x² + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )

d) x² + 8x + 7 = x² + x + 7x + 7 = x( x + 1 ) + 7( x + 1 ) = ( x + 1 )( x + 7 )

e) 3x² + 2x - 5 = 3x² - 3x + 5x - 5 = 3x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 3x + 5 )

f) 4x² - 3x - 1 = 4x² - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )

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

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

c, \(x^2-5x-6=x^2+x-6x-6=\left(x+1\right)\left(x-6\right)\)

d, \(x^2+8x+7=x^2+x+7x+7=\left(x+1\right)\left(x+7\right)\)

e, \(3x^2+2x-5=3x^2-3x+5x-5=\left(x-1\right)\left(3x+5\right)\)

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

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^