\(A=\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18\)

\(=\frac{1}{4}\left[\left(2x+1\right)\left(x+1\right)^2.4\left(2x+3\right)\right]-72\)

\(=\frac{1}{4}\left[\left(2x+1\right)\left(2x+3\right)\left(2x+2\right)^2\right]-72\)

\(=\frac{1}{4}\left[\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72\right]\)

Đặt: \(4x^2+8x+3=t\)

Ta có:  \(A=\frac{1}{4}\left[t^2+t-72\right]\)

\(=\frac{1}{4}\left[\left(t+9\right)\left(t-8\right)\right]\)

\(=\frac{1}{4}\left[\left(4x^2+8x+12\right)\left(4x^2+8x-5\right)\right]\)

\(=\left(x^2+2x+3\right)\left[4x^2+8x-5\right]\)

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

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

\(=\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]-4\)

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

Đặt \(12x^2+11x+2=a\)

Khi đó: \(B=a\left(a-3\right)-4\)

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

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

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

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

\(=x^4-2x^3+6x^2-8x+8\)

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

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

       \(3x^4-5x^3-18x^2-3x+5\)

\(=3x^4+x^3-x^2-6x^3-2x^2+2x-15x^2-5x+5\)

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

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

Bài này thật sự khó và hay đấy.

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

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

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

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

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

b)(x²+x+1)(x²+x+2)-12

Đặt t=x²+x+1

t(t+1)-12=t²+t-12

=(t-3)(t+4)=(x²+x+1-3)(x²+x+1+4)

=(x²+x-2)(x²+x+5)

=(x-1)(x+2)(x²+x+5)

c)(x²+8x+7)(x²+8x+15)+15

Đặt t=x²+8x+7 

t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+7+3)(x²+8x+7+15)

=(x²+8x+10)(x²+8x+22)

d)(x+2)(x+3)(x+4)(x+5)-24

=(x²+7x+10)(x²+7x+12)-24

Đặt t=x²+7x+10

t(t+2)-24=(t-4)(t+6)

=(x²+7x+10-4)(x²+7x+10+6)

=(x²+7x+6)(x²+7x+16)

=(x+1)(x+6)(x²+7x+16)

a/ Đặt x² + 4x + 8 = a

Thì đa thức ban đầu thành

a² + 3ax + 2x² = (a² + 2ax + x²) + (ax + x²)

= (a + x)² + x(a + x) = (a + x)(a + 2x)

b/ Đặt x² + x + 1 = a thì đa thức ban đầu thành

a(a + 1) - 12 = a² + a - 12 = (a² - 3a) + (4a - 12) 

= (a - 3)(a + 4)

đặt \(x^2+4x+8=a\)

=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)

          \(=\left(a+x\right)\left(a+2x\right)\)

b) ta có 

\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

đặt \(x^2+8x+11=a\)

=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)

         \(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)

         \(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)

khó thế

a)x⁴+2x³+5x²+4x-12

=(x⁴+2x³+x²)+(4x²+4x)-12

=(x²+x)²+4(x²+x)-12

Đặt t=x²+x

=t²+4t-12=(t-2)(t+6)

=(x²+x-2)(x²+x+6)

=(x-1)(x+2)(x²+x+6)

b)(x+1)(x+2)(x+3)(x+4)+1

=(x²+5x+4)(x²+5x+6)+1

Đặt x²+5x+4=t

t(t+2)+1=t²+2t+1

=(t+1)²=(x²+5x+4+1)²

=(x²+5x+5)²

c)(x+1)(x+3)(x+5)(x+7)+15

=(x²+8x+7)(x²+8x+15)+15

Đặt t=x²+8x+7

t(t+8)+15=(t+3)(t+5)

=(x²+8x+7+3)(x²+8x+7+5)

=(x²+8x+10)(x+2)(x+6)

d)(x+1)(x+2)(x+3)(x+4)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=x²+5x+4 

t(t+2)-24=(t-4)(t+6)

=(x²+5x+4-4)(x²+5x+4+6)

=x(x+5)(x²+5x+10)

Đặt x^2+2x=t =>3t^2-2t-1=3t^2-3t+t-1=3t(t-1)+(t-1)=(t-1)(3t+1)

=>(x^2+2x-1)(3x^2+6x+1)

(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 )

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

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

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

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

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

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

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