\(P=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}=\frac{3}{4}\sqrt{x^2+4}+\frac{\sqrt{x^2+4}}{4}+\frac{1}{\sqrt{x^2+4}}\)
\(\Rightarrow P\ge\frac{3}{4}\sqrt{x^2+4}+2\sqrt{\frac{\sqrt{x^2+4}}{4.\sqrt{x^2+4}}}\ge\frac{3}{4}\sqrt{4}+1=\frac{5}{2}\)
\(\Rightarrow P_{min}=\frac{5}{2}\) khi \(x=0\)
2/ \(\Delta=b^2-4ac\)
Ta có: \(c< b-a\)
\(\Rightarrow\Delta>b^2-4a\left(b-a\right)=b^2-4ab+4a^2=\left(b-2a\right)^2\ge0\)
\(\Rightarrow\Delta>0\Rightarrow\) pt luôn có 2 nghiệm pb