Question

So sánh:

a) \(\sqrt{2\sqrt{2}}\) và \(\sqrt{2}+1\)

b) \(\sqrt{12}-\sqrt{11}\) và \(\sqrt{11}-\sqrt{10}\)

Answer 1

a, \(\left(\sqrt{2\sqrt{2}}\right)^2=2\sqrt{2}< 2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)

=> \(2\sqrt{2}< \sqrt{2}+1\)( vì \(2\sqrt{2}>0,\sqrt{2}+1>0\))

b, \(1=\left(\sqrt{12}-\sqrt{11}\right)\left(\sqrt{12}+\sqrt{11}\right)\)

=> \(\sqrt{12}-\sqrt{11}=\frac{1}{\sqrt{12}+\sqrt{11}}\)

Tương tự: \(\sqrt{11}-\sqrt{10}=\frac{1}{\sqrt{11}+\sqrt{10}}\)

Do \(\sqrt{12}+\sqrt{11}>\sqrt{11}+\sqrt{10}\)<=> \(\sqrt{12}-\sqrt{11}=\frac{1}{\sqrt{12}+\sqrt{11}}< \frac{1}{\sqrt{11}+\sqrt{10}}=\sqrt{11}-\sqrt{10}\)

=> \(\sqrt{12}-\sqrt{11}< \sqrt{11}-\sqrt{10}\)