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

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

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

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

\(e,4x^2+4x-15=\left(4x^2+4x+1\right)-16=\left(2x+1\right)^2-4^2=\left(2x+5\right)\left(2x-3\right)\)

\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)

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

Phân tích à :v

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

b) x⁴ + x³ + x² - 1 = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

c) x² + 2xy + y² - 2x - 2y + 1 = ( x² + 2xy + y² ) - ( 2x + 2y ) + 1 = ( x + y )² - 2( x + y ) + 1² = ( x + y - 1 )²

d) x²y² + 1 - x² - y² = ( x²y² - x² ) - ( y² - 1 ) = x²( y² - 1 ) - ( y² - 1 ) = ( y² - 1 )( x² - 1 ) = ( y - 1 )( y + 1 )( x - 1 )( x + 1 )

e) 4x² + 4x - 15 = ( 4x² + 4x + 1 ) - 16 = ( 2x + 1 )² - 4² = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )

g) 3x² - 7x + 2 = 3x² - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )

h) 4x² - 5x + 1 = 4x² - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )

Chứng minh

a) \(2\equiv-1\left(mod3\right)\)

\(\Rightarrow2^{1000}\equiv\left(-1\right)^{1000}\equiv1\left(mod3\right)\Rightarrow2^{1000}-1\equiv0\left(mod3\right)\Rightarrowđpcm\)

b) \(19\equiv-1\left(mod20\right)\)

\(\Rightarrow19^{45}\equiv\left(-1\right)^{45}\equiv1\left(mod20\right);19^{30}\equiv\left(-1\right)^{30}\equiv1\left(mod20\right)\)

\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod20\right)\Rightarrowđpcm\)

a) x³ - 6x² + 11x - 6

= ( x³ - 2x² ) - ( 4x² - 8x ) + ( 3x - 6 )

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

= ( x - 2 )( x² - 4x + 3 )

= ( x - 2 )( x² - x - 3x + 3 )

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

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

b) x³ - 6x² - 9x + 14

= ( x³ - x² ) - ( 5x² - 5x ) - ( 14x - 14 )

= x²( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )

= ( x - 1 )( x² - 5x - 14 )

= ( x - 1 )( x² + 2x - 7x - 14 )

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

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

c) x³ + 6x² + 11x + 6

= ( x³ + 2x² ) + ( 4x² + 8x ) + ( 3x + 6 )

= x²( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )

= ( x + 2 )( x² + 4x + 3 )

= ( x + 2 )( x² + x + 3x + 3 )

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

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

e) x⁶ - 9x³ + 8

Đặt t = x³

bthuc <=> t² - 9t + 8 

            = t² - t - 8t + 8

            = t( t - 1 ) - 8( t - 1 )

            = ( t - 1 )( t - 8 )

            = ( x³ - 1 )( x³ - 8 )

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

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

\(5x^2+5xy-x-y\)

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

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

Sửa đề : a, \(x\left(x-2\right)^2-\left(x-2\right)\left(x^2+2x+4\right)\)

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

\(=-4x^2+4x+8\)

x( x - 2 )² - ( x - 2 )( x² + 2x + 4 ) < bỏ cái ² đi nhá :v >

= x( x² - 4x + 4 ) - ( x³ - 8 )

= x³ - 4x² + 4x - x³ + 8

= -4x² + 4x + 8

a) \(x^{m+2}-2x^m=x^m\left(x^2-2\right)\)

b) \(x^{k+1}-x^{k+2}=x^{k+1}\left(1-x\right)\)

a) x^m+2 - 2x^m = x^m.x² + 2.x^m = x^m( x² - 2 ) = x^m( x - √2 )( x + √2 )

b) x^k+1 - x^k+2 = x^k+1 - x^k+1.x = x^k+1( 1 - x )

PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi 

Giải ( suỵt :), đừng ai nhìn thấy ... :v 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = {5;-10}

sao lại "vô lí" vậy bạn 

\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)

Ta có: \(x^3+9x^2+26x+24\)

\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)

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

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

\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)

\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

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

Đến đây là PT tích r còn gì, \(x\in\left\{5;-10;-\sqrt{3}\right\}\)

:v ez mà :) Sửa đề : tìm x 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = 5 ; x = -10