\(15-2x-x^2=0\\ \Leftrightarrow x^2+2x-15=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(5x-15\right)=0\\ \Leftrightarrow x\left(x-3\right)+5\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy...
\(15-2x-x^2=0\\ \Leftrightarrow x^2+2x-15=0\\ \Leftrightarrow x^2+2x+1=16\\ \Leftrightarrow\left(x+1\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
15 - 2x - x2 = 0
x2 + 2x - 15 = 0
x2 - 32 + 2x - 6 =0
( x - 3)( x +3) + 2( x - 3) =0
( x - 3)( x + 5) = 0
Vậy , x nhận hai giá trị là : 3 ; -5
