Ta có: \(\left(x^2+4x+3\right)\left(x^2+6x+8\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x+2\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)(1)
Ta có: \(1\cdot2\cdot3\cdot4=24\)(2)
Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}x+1=1\\x+2=2\\x+3=3\\x+4=4\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Sai từ chỗ (1)
(1)\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)(2)
Đặt \(y=x^2+5x+4=\left(x+1\right)\left(x+4\right)\)
\(\left(1\right)\Leftrightarrow y^2+2y-24=0\)
\(\Leftrightarrow\left(y-4\right)\left(y+6\right)=0\Rightarrow\left[{}\begin{matrix}y=4\\y=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\end{matrix}\right.\)
Vậy x=0 hoặc x=-5
Ta cos (x2+4x+3)(x2+6x+8)=24
<=>(x+1)(x+2)(x+3)(x+4)-24=0
<=>(x+1)(x+4)(x+2)(x+3)-24=0
<=>(x2+5x+4)(x2+5x+6)-24=0
Đặt a=x2+5x+5 =>(a-1)(a+1)-24=0
<=>a2-1-24=0
<=>a2-25=0
<=>(a-5)(a+5)=0
<=>a=5
hoặc a=-5
<=>x2+5x+5=5
Hoặc x2+5x+5=-5
<=>x(x-5)=0<=>\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Hoặc x2+5x+10=0(vô lí)
Vậy tập nghiệm của pt là\(\left\{0,5\right\}\)