Đặt \(x^2-2x+4=a\)

Khi đó \(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8=\left(a-1\right)\left(a+1\right)-8\)

                                                                                    \(=a^2-1-8\)

                                                                                    \(=a^2-9\)

                                                                                      \(=\left(a-3\right)\left(a+3\right)\)

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

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

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

Đặt \(2x^2-x-2=t\)

Ta có:

\(A=\left(t+3\right)\left(t-3\right)+8\)

\(A=t^2-9+8\)

\(A=\left(t-1\right)\left(t+1\right)\)

Thay vào ta được:

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

a, \(x^3-2x-y^3+2y\) (sửa đề)

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

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

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

b, \(\left(x-y\right)\left(x+y\right)-4zx+4yz\)

\(=\left(x-y\right)\left(x+y\right)-\left(4zx-4yz\right)\)

\(=\left(x-y\right)\left(x+y\right)-4z\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-4z\right)\)

Bạn xem lại đề câu a giúp mình nha!

\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)

\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)

\(=x^8+12x^5-2x^4+36x^2-12x-99\)

\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)

\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)

\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)

đặt y=x²+4x+8 ta được

y²+3xy+2x²=y²+xy+2xy+2x²=y(y+x)+2x(y+x)

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

thay y=x²+4x+8 ta được

(x²+5x+8)(x²+7x+8)

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

=(x^2+5x+8)(x^2+6x+8)

(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2

dat x^2+4x+8=y

ta co:y^2+3xy+2x^2

=y^2+xy+2xy+2x^2

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

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

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

=(x^2+6x+8)(x^2+5x+8)

KL:......................

B

(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 )

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

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

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

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

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

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

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

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

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

\(a.\) \(ax^2-a^2x-x+a\)

\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(ax-1\right)\left(x-a\right)\)

\(b.\) \(18x^3-12x^2+2x\)

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

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

\(c.\) \(x^3-5x^2-4x+20\)

\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)

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

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

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

\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)

\(=x^2+15x+7x+105+15\)

\(=x^2+22x+120\)

\(=\left(x+10\right)\left(x+12\right)\)