Question

x^2+5y^2+2x+6y+34

Answer 1

\(\)\(x^2+5y^2+2x+6y+34\)
\(=\left(x^2+2x.1+1\right)+\left[\left(\sqrt{5}y\right)^2+2.\sqrt{5}x.\dfrac{3}{\sqrt{5}}+\left(\dfrac{3}{\sqrt{5}}\right)^2\right]+\dfrac{156}{5}\)
\(=\left(x+1\right)^2+\left(\sqrt{5}y+\dfrac{3}{\sqrt{5}}\right)^2+\dfrac{156}{5}\)
Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2\text{≥ 0}\\\left(\sqrt{5}y+\dfrac{3}{\sqrt{5}}\right)^2\text{≥ 0}\end{matrix}\right.\)
⇒  \(\left(x+1\right)^2+\left(\sqrt{5}y+\dfrac{3}{\sqrt{5}}\right)^2+\dfrac{156}{5}\text{ }\text{≥}\dfrac{156}{5}\)
⇒ đa thức vô nghiệm