$Ta có:$
$\frac{89}{45} =2-\frac{1}{45} $
$và \frac{1}{45} =\frac{2}{90} =\frac{2}{\sqrt[]{8100} } $
$Vì \sqrt[]{2010} <\sqrt[]{8100} ⇔\frac{2}{\sqrt[]{2010}} >\frac{2}{\sqrt[]{8100}} $
$⇔ \frac{2}{\sqrt[]{2010}}>\frac{2}{90}$
$⇔\frac{2}{\sqrt[]{2010}}>\frac{1}{45}$
$⇔2-\frac{2}{\sqrt[]{2010}}<2-\frac{1}{45}$
$⇔2-\frac{2}{\sqrt[]{2010}}<\frac{89}{45}.(đpcm)$
