\(x^2+x^2-5x+3x-15=9\)
\(2x^2-2x-15-9=0\)
\(2x^2-2x-24=0\)
\(x^2-x-12=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=4\\x_2=-3\end{matrix}\right.\)
\(x^2+\left(x+3\right)\left(x-5\right)=9\)
\(\Leftrightarrow x^2+x^2-5x+3x-15-9=0\)
\(\Leftrightarrow2x^2-2x-24=0\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.2.\left(-24\right)=196>0\Rightarrow\sqrt{\Delta}=14\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{2+14}{4}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{2-14}{4}=-3\end{matrix}\right.\)
Vậy \(S=\left\{4;-3\right\}\)
\(x^2+\left(x+3\right)\left(x-5\right)=9\)
\(x^2+x^2-5x+3x-15=9\)
\(2x^2-2x-24=0\)
\(x^2-x-12=0\)
\(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{49}{4}\)
\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{49}{4}=\left(\dfrac{7}{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}+\dfrac{1}{2}=4\\x=\dfrac{-7}{2}+\dfrac{1}{2}=-3\end{matrix}\right.\)
