Bài 2:

a: \(=\left(4x^2+6x+2x+3\right)\left(x^2+2x+1\right)-18\)

\(=\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18\)

Đặt x^2+2x=a

\(A=\left(4a+3\right)\left(a+1\right)-18\)

\(=4a^2+7a-15\)

\(=4a^2+12a-5a-15=\left(a+3\right)\left(4a-5\right)\)

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

\(=\left(x^2+2x+3\right)\left(4x^2+10x-2x-5\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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)

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)

khos quas

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

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

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

\(4,=2x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(2x-6\right)=\left(x-y\right)2\left(x-3\right)\)

\(5,=4x\left(x+3\right)\)

1) x² . (x+1) + 2x . ( x+1) = (x² + 2x) . (x+1)

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

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

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

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

4) 2x (x-y) + 6 (y-x) = 2x (x -y) - 6 (x-y) 

                               = (2x-6) . (x-y)

5) 4x² + 12x = x . (4x+12)

mik mới hok lớp 7 nên làm thử nếu đúng thì các bạn nhớ ủng hộ mik nha còn nếu sai mong các bạn thông cảm

Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+10=y\)

\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)

\(A=y^2+2y+1-25\)

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

\(A=\left(y+1-5\right)\left(y+1+5\right)\)

\(A=\left(y-4\right)\left(y+6\right)\)

\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

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

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

Đặt \(12x^2+11x-1=a\)

\(\Rightarrow B=a.\left(a+3\right)-4\)

\(B=a^2+3a-4\)

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

\(B=a.\left(a-1\right)+4.\left(a-1\right)\)

\(B=\left(a-1\right)\left(a+4\right)\)

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

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

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

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

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

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

Đặt \(12x^2+11x=t\), ta có:

\(\left(t+2\right)\left(t-1\right)-4\)

\(=t^2-t+2t-2-4=t^2+t-6\)

\(=t^2-2t+3t-6\)

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

Thay \(t=12x^2+11x\), ta được:

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

Đs...