\(\sqrt{5}-\dfrac{5}{2}=\dfrac{2\sqrt{5}-5}{2}=\dfrac{\sqrt{20}-\sqrt{25}}{2}\)
ta có \(20< 25\ge\Rightarrow\sqrt{20}-\sqrt{25}< 0\Rightarrow\dfrac{\sqrt{20}-\sqrt{25}}{2}< 0\)
vậy \(\sqrt{5}-\dfrac{5}{2}< 0\)
Ta có:
\(\sqrt{5}\) = \(\dfrac{5}{\sqrt{5}}\) < \(\dfrac{5}{\sqrt{4}}\) = \(\dfrac{5}{2}\)
<=> \(\sqrt{5}\) < \(\dfrac{5}{2}\)
<=> \(\sqrt{5}\) - \(\dfrac{5}{2}\) < 0
Vậy : \(\sqrt{5}\) - \(\dfrac{5}{2}\) < 0
Vì : \(\sqrt{5}-\dfrac{5}{2}=\sqrt{5}-2.5=\sqrt{5}-\sqrt{6.25}< 0\)
nên \(\sqrt{5}-\dfrac{5}{2}< 0\)
Vậy : \(\sqrt{5}-\dfrac{5}{2}< 0\)