\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)-42=0\)
Đặt \(a=x^2+x\) ta có:
\(a\left(a+1\right)-42=0\)
\(\Leftrightarrow a^2+a-42=0\)
\(\Leftrightarrow a^2-6a+7a-42=0\)
\(\Leftrightarrow a\left(a-6\right)+7\left(a-6\right)=0\)
\(\Leftrightarrow\left(a-6\right)\left(a+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)
* \(a=6\Leftrightarrow x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
* \(a=-7\Leftrightarrow x^2+x=-7\Leftrightarrow x^2+x+7=0\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}=0\left(l\right)\)
Vậy: S = {2;-3)
\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)
Đặt : \(x^2+x=d\) thì phương trình trở thành :
\(d\left(d+1\right)=42\)
\(\Leftrightarrow d^2+d-42=0\)
\(\Leftrightarrow\left(d-6\right)\left(d+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}d-6=0\\d+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}d=6\\d=-7\end{matrix}\right.\)
Với \(d=6\) thì :
\(x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Với \(d=-7\) thì :
\(x^2+x=-7\)
\(\Leftrightarrow x^2+x+7=0\)
\(\Rightarrow\) Phương trình vô nghiệm ( Cái này bạn tự chứng minh nó vô nghiệm nhé )
Vậy \(S=\left\{-3;2\right\}\)
