\(\left(x^2+2x\right)^2-14\left(x^2+2x\right)-15=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+\left(x^2+2x\right)-15\left(x^2+2x\right)-15=0\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x+1\right)-15\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-3\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-5\end{matrix}\right.\)
Vậy...
Bạn tham khảo nhé!
Ta có: \(\left(x^2+2x\right)^2-14\left(x^2+2x\right)-15=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+\left(x^2+2x\right)-15\left(x^2+2x\right)-15=0\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x+1\right)-15\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\\x=3\end{matrix}\right.\)
Vậy: S={-1;-5;3}
