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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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
= \(\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
= \(\left(x^2+4x+8\right)\left(x^2+4x+8+x\right)+2x\left(x^2+4x+8+x\right)\)
= \(\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)
= \(\left(x^2+2x+4x+8\right)\left(x^2+5x+8\right)\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(S=\left\{1;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x^2+8x+11=a\)
=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)
\(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)
(x+2)(x+4)(x^2+5x+8)
#)Giải :
Đặt \(x^2+4x+8=k\)
Ta có :\(k^2+3xk+2x^2=k^2+2xk+xk+2x^2=k\left(k+2x\right)+x\left(k+2x\right)=\left(k+x\right)\left(k+2x\right)\)
\(\Rightarrow\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8+x\right)\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)