Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

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

b/ \(=y^4+\left(-2x^2-34\right)y^2+32xy+x^4-34x^2+225\)

câu này bn coi lại đc k , mk k lm ra 

\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

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

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

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

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

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

\(4x^2-9y^2+4x-6y\)

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

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

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

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

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~

(1 + x²)² - 4x(1 - x²)

= (1 + x²)(1 + x²) - 4x(1 - x²)

= (1 + x² - 4x)(1 + x² - 1 + x²)

= 2x²(x² - 4x + 1)

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

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

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

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

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

\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)

\(=x^4+8x^3+11x^2-44x+24\)

\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)

\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)

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

\(=x^2\left(x-3\right)-\left(4x-12\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

x²-4x+3

=x.x-x-3x-3

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

=(x-1)(x-3)

x² - 4x + 3

= x² - x - 3x + 3 ( dùng phương pháp tách nhân tử )

= ( x² - x ) - ( 3x - 3 ) ( vì đằng trc có đấu " - " nên ta đổi đáu trong ngoặc )

= x( x - 1 ) - 3(x - 1 ) ( đặt nhân tử chung )

= ( x - 1 ) ( x - 3 )

hok tốt !