\(\left(x^2+x\right)^2+4.\left(x^2+x\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+4.\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right).\left(x^2+x+4-12\right)=0\)
\(\Leftrightarrow\left(x^2+x\right).\left(x^2+x-8\right)=0\)
\(\Leftrightarrow x.\left(x+1\right).\left(x^2+x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x^2+x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-3,372281323\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{0;-1;-3,372281323\right\}.\)
Chúc bạn học tốt!
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+6\right)-2\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x^2+2x-x-2\right)\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{23}{4}\right)=0\)
\(\Leftrightarrow\left[x\left(x+2\right)-\left(x+2\right)\right]\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\) (vì \(\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]>0\))
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy S = {-2;1}
