Ta có: \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\)\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+2.2.\left(x^2+x\right)+4-16=0\)
\(\Leftrightarrow\left[\left(x^2+x\right)+2\right]^2-4^2=0\)
\(\Leftrightarrow\left(x^2+x+2-2\right)\left(x^2+x+2+2\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+4\right)=0\)
Vì: \(x^2+x+4=x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{15}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
\(\Rightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy x = 0; x = -1.
Chúc pạn hok tốt!!! Mk k chắc lắm đâu!!!
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+2\right)^2-16=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+6\right)=0\)
+)\(x^2+x-2=0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{9}{4}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
+)\(x^2+x+6=0\Leftrightarrow x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\)(vô nghiệm)
Vậy S=\(\left\{1;-2\right\}\)
