\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15=0\)
Đặt \(x^2+x=a\)
\(pt\Leftrightarrow a^2-2a-15=0\)
\(\Leftrightarrow a^2+5a-3a-15=0\)
\(\Leftrightarrow a\left(a+5\right)-3\left(a+5\right)=0\)
\(\Leftrightarrow\left(a+5\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-5\\a=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x=-5\\x^2+x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{19}{4}=0\\x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=-\dfrac{19}{4}\left(loai\right)\\\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{\pm\sqrt{13}}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\left\{\dfrac{\sqrt{13}-1}{2};\dfrac{-\sqrt{13}-1}{2}\right\}\end{matrix}\right.\)
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