Question

x(x+1)(x+2)(x+3)=24

Answer 1

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\) (1)

\(\Leftrightarrow\left(x^2+x\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\left(x^3+2x^2+x^2+2x\right)\left(x+3\right)=24\)

\(\Leftrightarrow\left(x^3+3x^2+2x\right)\left(x+3\right)=24\)

\(\Leftrightarrow x^4+3x^3+3x^3+9x^2+2x^2+6x=24\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x=24\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x-24=0\)

\(\Leftrightarrow x^4-x^3+7x^3-7x^2+18x^2-18x+24x-24=0\)

\(\Leftrightarrow x^3\left(x-1\right)+7x^2\left(x-1\right)+18x\left(x-1\right)+24\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+7x^2+18x+24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+3x^2+12x+6x+24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[x^2\left(x+4\right)+3x\left(x+4\right)+6\left(x+4\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x^2+3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+4=0\\x^2+3x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\\x\notin R\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{-4;1\right\}\)

Answer 2

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Rightarrow\left[x\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]-24=0\)

\(\Rightarrow\left(x^2+3x\right)\left(x^2+2x+x+2\right)-24=0\)

\(\Rightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt \(x^2+3x=t\Rightarrow x^2+3x+2=t+2\)

\(\Rightarrow t.\left(t+2\right)-24=0\)

\(\Rightarrow t^2+2t-24=0\)

\(\Rightarrow t^2-4t+6t-24=0\)

\(\Rightarrow t.\left(t-4\right)+6.\left(t-4\right)=0\)

\(\Rightarrow\left(t-4\right).\left(t+6\right)=0\)(1)

Vì \(x^2+3x=t\) nên

\(\left(1\right)=\left(x^2+3x-4\right).\left(x^2+3x+6\right)=0\)

\(\Rightarrow\left(x^2-x+4x-4\right).\left(x^2+3x+6\right)=0\)

\(\Rightarrow\left(x-1\right).\left(x+4\right)\left(x^2+3x+6\right)=0\)

Ta có:

\(x^2+3x+6=x^2+1,5x+1,5x+2,25+3,75\)

\(=\left(x+1,5\right)^2+3,75\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x+1,5\right)^2\ge0\Rightarrow\left(x+1,5\right)^2+3,75\ge3,75>0\)

\(\Rightarrow\left(x-1\right).\left(x+4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy......

Chúc bạn học tốt!!!

Answer 3

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)

Đặt t = \(x^2+3x+1\)

\(=>t^2-1=24\)

\(=>t^2=25\)

\(=>\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

Khi \(t=5=>x^2+3x+1=5\)

\(\Leftrightarrow x^2+4x-x-4=0\)

\(\Leftrightarrow x\left(x+4\right)-\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(=>\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)(1)

Khi \(t=-5=>x^2+3x+6=0\)

\(=>x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{15}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{9}{4}\)

Vậy không có giá trị x thoả mãn .(2)

Từ (1) và (2) :

Vậy tập nghiệm của pt là : \(S=\left\{1;-4\right\}\)