\(x^2+5x+5=0\)
\(\Leftrightarrow x^2+x+5x+5=0\)
\(\Leftrightarrow\left(x^2+x\right)+\left(5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
Vậy ....
\(\Leftrightarrow x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}=0\) ( vô lý )
Ta có :
\(\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}>0\forall x\)
Vậy pt vô nghiệm
\(x^2+5x+5=0\)
\(\Leftrightarrow x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{5}{2}=\dfrac{-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-5}{2}\\x=\dfrac{-\sqrt{5}-5}{2}\end{matrix}\right.\)
