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

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

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

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

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

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

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

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

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

\(x^4+x^3+2x^2+x+1\)

\(=x^4+x^3+x^2+x^2+x+1\)

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

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

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

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\\ =4\left(x^2+16x+60\right)\left(x^2+17x+60\right)-3x^2\)

Đặt \(x^2+16x+60=a\)

\(=4a\left(a+x\right)-3x^2\\ =4a^2+4ax-3x^2\\ =\left(2a-x\right)\left(2a+3x\right)\\ =\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\\ =\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

Đặt 

a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

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

a: \(5x^2-4\left(x^2-2x+1\right)-5\)

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

b: \(9x^2+6x-4y^2+4y\)

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

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

a, \(x^2-4x+3=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

TH1 : x = 3 ; TH2 : x = 1

b, \(2x^2-3x-2=0\Leftrightarrow\left(x-2\right)\left(x+\frac{1}{2}\right)=0\)

TH1 : x = 2 ; TH2 : x = -1/2 

c, Đặt \(x^2=t\left(t\ge0\right)\)

\(t^2+2t-8=0\Leftrightarrow\left(t-2\right)\left(t+4\right)=0\)

TH1 : t  = 2 ; TH2 : t = -4 

Tương tự ... 

1a) 

x² - 4x + 3 = x² - x - 3x + 3 

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

                  = ( x - 1 )( x - 3 )

2c) 

2x² - 3x - 2 = 2x² + x - 4x - 2 

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

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

3e)

x⁴ + 2x² - 8 (*)

Đặt t = x²

(*) <=> t² + 2t - 8

       = t² - 2t + 4t - 8 

       = t( t - 2 ) + 4( t - 2 )

       = ( t - 2 )( t + 4 )

       = ( x² - 2 )( x² + 4 )

4b) x² + 4x - 12 = x² - 2x + 6x - 12

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

                          = ( x - 2 )( x + 6 )

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

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

f) x² - 2xy - 3y² = ( x² - 2xy + y² ) - 4y²

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

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

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

Phân tích đa thức thành nhân tử à .... sr, giải PT mất tiuu 

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

\(= (x+4)^2(x^2-1)-(x^2-1)=[(x+4)^2-1](x^2-1)\)

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

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

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

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

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

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

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

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

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

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

Tham Khảo

2x² + x - 6

= 2x² + 4x - 3x - 6

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

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