Ta có :

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

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

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

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

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

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

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

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

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

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

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

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

= (x - 5 + 2x - 1)² - (2x - 6)²

= (3x - 6)² - (2x - 6)²

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

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

= [ ( x - 5 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )² ] - 2²( x - 3 )²

= ( x - 5 + 2x - 1 )² - ( 2x - 6 )²

= ( 3x - 6 )² - ( 2x - 6 )²

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

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

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

\(=\left(x-5+2x-1\right)^2-\left(2x-6\right)^2\)

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

\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)

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

\(=\left(5x-12\right)x\)

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

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

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

Lời giải:

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

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

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

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

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

`(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)`

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

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

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

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

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

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

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

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

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

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

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

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

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

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

=.= hok tốt!!

Tacó:

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

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

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

Dat \(a=x^4+x^2\)

\(A=a\left(a-2\right)+1=\left(a-1\right)^2\)

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

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

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

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

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

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

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

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

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

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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