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:......................

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

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

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

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

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

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

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

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

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

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

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

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

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

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{1;3\right\}\)

\(d,x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(x-2\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

d: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

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

=>x=1 hoặc x=3

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