#)Giải :

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

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

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

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

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

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

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

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

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

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

cảm ơn nha!

a) x² + 4x + 3

= x² + 3x + x +3

= ( x² + 3 ) + ( x + 3 )

= x ( x + 3 ) + ( x + 3 )

= ( x + 3 ) ( x + 1 )

b) 4x² - 4x - 3

= 4x² + 2x - 6x - 3

= ( 4x² + 2x ) - ( 6x + 3 )

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

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

c) x² - x - 12

= x² + 3x - 4x - 12

= ( x² + 3x ) - ( 4x + 12 )

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

= ( x + 3 ) ( x - 4 )

d) 4x⁴ - 4x²y² - 8y⁴

= 4 ( x⁴ - x²y² - 2y⁴ )

Hk tốt

cảm ơn bạn

a) \(x^2+4x+3\)

= \(x^2+x+3x+3\)

= \(x\left(x+1\right)+3\left(x+1\right)\)

= \(\left(x+1\right)\left(x+3\right)\)

b) \(4x^2-4x-3\)

= \(4x^2+2x-6x-3\)

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

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

c)\(x^2-x-12\)

= \(x^2-4x+3x-12\)

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

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

câu d la 4x^2.y^2 phai ko

\(4x^4-4x^2y^2-8y^4\)

= \(4x^4-8x^2y^2+4x^2y^2-8y^4\)

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

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

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

Kết bạn với mình nha

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: \(xy-4y-5x+20\)

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Lời giải:
a.

$x^4+10x^3+26x^2+10x+1$

$=(x^4+10x^3+25x^2)+x^2+10x+1$

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

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

$=(x^2+4x+1)(x^2+6x+1)$

b.

$x^4+x^3-4x^2+x+1$

$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$

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

$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$

$=(x-1)(x^3+2x^2-2x-1)$

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

$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$

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

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

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

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

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

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

\(x^5+x+1\)

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

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

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

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

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

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

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

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

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

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

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

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

c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)

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

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

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

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