x²(x²-6)-x²+9

<=>(x⁴-6x²+9)-x²

<=>(x²-3)²-x²

<=>(x²-3-x)(x²-3+x)

Phân tích đa thức thành nhân tử

(x+3)(x−6)+x²−9

Tk                                      nha !

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

\(=x^2-3x-18+x^2-9\)

\(=2x^2-3x-27\)

\(=\left(2x^2+6x\right)-\left(9x+27\right)\)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

Rút gọn thôi chứ phân tích sao được ._.

( x - 3 )² - ( 4x + 5 )² - 9( x + 1 )² - 6( x - 3 )( x + 1 )

= x² - 6x + 9 - ( 16x² + 40x + 25 ) - 9( x² + 2x + 1 ) - 6( x² - 2x - 3 )

= x² - 6x + 9 - 16x² - 40x - 25 - 9x² - 18x - 9 - 6x² + 12x + 18

= -30x² - 52x - 7

Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))

Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)

\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)

\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)

\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)

\(=\left(4x+7\right)\left(12x+17\right)\)

Ta có : (x³ + 3x + 1)(x³ + 3x + 2) - 6

= (x³ + 3x + 1,5 - 0,5)(x³ + 3x + 1,5 + 0,5) - 6

= (x³ + 3x + 1,5)² - 0,5² - 6

= (x³ + 3x + 1,5)² - 6,25

= (x³ + 3x + 1,5 - 2,5) (x³ + 3x + 1,5 + 2,5)

= (x³ + 3x - 1) (x³ + 3x + 3)

`(x+3)^4+(x+5)^4-2`

`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`

`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`

`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`

`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`

`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`

`=(x+4)(2x^3+24x^2+108x+176)`

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

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

\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)

\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)

\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)

\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)

\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)

\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)

\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)

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

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

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

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

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
 

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)

\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

\(\left(x^2-10x+20\right)^2\)

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

\(=\left[\left(x-5\right)^2-5\right]^2\)

\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)

      \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+18\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)-16\)

\(=\left(x^2+10x+20\right)^2-16+16=\left(x^2+10x+20\right)^2\)

Chúc bạn học tốt.

      \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(\Rightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+6\right)\left(x+8\right)\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left[\left(x^2+10x+16\right)+8\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+4^2\)

\(\Rightarrow\left(x^2+10x+20\right)^2\)