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) \(x^2+12x+35\)

\(=x^2+5x+7x+35\)

\(=\left(x^2+5x\right)+\left(7x+35\right)\)

\(=x\left(x+5\right)+7\left(x+5\right)\)

\(=\left(x+5\right)\left(x+7\right)\)

b)\(x^2-x-56\)

\(=x^2+7x-8x-56\)

\(=\left(x^2+7x\right)-\left(8x+56\right)\)

\(=x\left(x+7\right)-8\left(x+7\right)\)

\(=\left(x+7\right)\left(x-8\right)\)

c)\(5x^2-x-4\)

\(=5x^2-5x+4x-4\)

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

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

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

TL:

a)\(x^2+5x+7x+35\) 

 =\(x\left(x+5\right)+7\left(x+5\right)\) 

=\(\left(x+7\right)\left(x+5\right)\) 

b) \(x^2-x-56\)

  =\(x^2+7x-8x-56\) 

=\(x\left(x+7\right)-8\left(x+7\right)\) 

=\(\left(x-8\right)\left(x+7\right)\) 

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

=\(\left(2x^2+1\right)^2-4x^2\) 

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

.......................(tự lm)

hc tốt

\(4\times^4+81\)

\(=\left(2\times^2\right)^2+36\times^2+9^2-36\times^2\)

\(=\left(2\times^2+9\right)^2-36\times^2\)

\(=\left(2\times^2+9+36\times\right)\left(2\times^2+9-36\times\right)\)

1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)

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

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

3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)

a) x²(x-3)-4x+12

=x²(x-3)-4(x-3)

=(x-3)(x²-4)

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

b) 2a(x+y)-x-y

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

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

c) 2x-4+5x²-10x

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

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

d) 5x²-12x-7x+14

=5x²-19x+14

e) xy-y²-3x+3y

=y(x-y)-3(x-y)

=(x-y)(y-3)

#H

Trả lời:

1, x² ( x - 3 ) - 4x + 12 = x² ( x - 3 ) - 4 ( x - 3 ) = ( x - 3 )( x² - 4 ) = ( x - 3 )( x - 2 )( x + 2 )

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

3, 2x - 4 + 5x² - 10x = ( 2x - 4 ) + ( 5x² - 10x ) = 2 ( x - 2 ) + 5x ( x - 2 ) = ( x - 2 )( 2 + 5x )

4, Sửa đề: 6x² - 12x - 7x + 14 = ( 6x² - 12x ) - ( 7x - 14 ) = 6x ( x - 2 ) - 7 ( x - 2 ) = ( x - 2 )( 6x - 7 )

5, xy - y² - 3x + 3y = ( xy - y² ) - ( 3x - 3y ) = y ( x - y ) - 3 ( x - y ) = ( x + y )( y - 3 )

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

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

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

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

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

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

\(x^4+2x^2-24\)

\(=x^4+2x^2+1-25\)

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

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

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

a) `x^4+2x^3-4x-4`

`=(x^4-4)+(2x^3-4x)`

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

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

b) `x^3-4x^2+12x-27`

`=(x^3-27)-(4x^2-12x)`

`=(x-3)(x^2+3x+9)-4x(x-3)`

`=(x-3)(x^2+3x+9-4x)`

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

c) `xy-4y-5x+20`

`=y(x-4)-5(x-4)`

`=(y-5)(x-4)`

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

b) Ta có: \(x^3-4x^2+12x-27\)

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

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

c) Ta có: \(xy-4y-5x+20\)

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

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

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

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

bạn ktra lại đề

b)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

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

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

Ủa pạn có thể giải ại cái bước thứ 2 đc ko ạk

tự làm

Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết

a) \(x^3-4x^2-12x+27\)

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

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

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

b) \(x^3-3x^2-4x+12\)

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

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

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

a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)

b) \(6x-9-x^2=-\left(x-3\right)^2\)

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

d) \(\left(x-y\right)^2-\left(x+y\right)^2\)

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

\(=2x.\left(-2y\right)=-4xy\)

e) \(\left(2x-1\right)^2-\left(x-1\right)^2\)

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

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

g) \(x^2+5x-6\)

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

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

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