\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)
Đặt: \(x^2+x=v\) ta được pt mới:
\(v\left(v+1\right)=42\)
\(\Leftrightarrow\left(v-6\right)\left(v+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}v=6\\v=-7\end{matrix}\right.\)
Khi \(v=6\) ta có:
\(x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Khi \(v=-7\) ta có:
\(x^2+x+7=0\)
\(\Leftrightarrow x^2+2x.\frac{1}{2}+\frac{1}{4}+\frac{27}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{27}{4}=0\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\forall x\)
\(\Rightarrow\) Vô nghiệm
Vậy ..........
\(x\left(x+1\right)\left(x^2+x+1\right)=42\\ \left(x^2+x\right)\left(x^2+x+1\right)=42\)
Đặt \(y=x^2+x\) ta có:
\(y\left(y+1\right)=42\\\Leftrightarrow y^2+x-42=0\\ \Leftrightarrow y^2-6y+7y-42=0\\\Leftrightarrow y\left(y-6\right)+7\left(y-6\right)=0\\\Leftrightarrow \left(y+7\right)\left(y-6\right)=0\\\Leftrightarrow \left[{}\begin{matrix}y+7=0\\y-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-7\\y=6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\end{matrix}\right.\Rightarrow x^2+x-6=0\left(Vix^2+x+7=0lasai\right)\\ \Leftrightarrow x^2-2x+3x-6=0\\\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)Vậy tập nghiệm của phương trình là \(S=\left\{-3;2\right\}\)
