Question

so sánh :

\(2+\sqrt{2}\) và \(3-\sqrt{3}\)

Answer 1

Lời giải:

$2+\sqrt{2}>2$

$3-\sqrt{3}<3-\sqrt{1}=2$

$\Rightarrow 2+\sqrt{2}>3-\sqrt{3}$

Answer 2

Ta có: \(\left(2+\sqrt{2}\right)^2=4+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2=4+4\sqrt{2}+2=6+4\sqrt{2}\)

\(=6+\sqrt{32}\)

\(\left(3-\sqrt{3}\right)^2=3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=9-6\sqrt{3}+3=12-6\sqrt{3}\)

\(=12-\sqrt{108}\)

Ta có: \(\sqrt{32}< \sqrt{108}\)(vì 32<108)

\(\Rightarrow\sqrt{32}>-\sqrt{108}\)

\(\Rightarrow\sqrt{32}+6>-\sqrt{108}+12\)

hay \(2+\sqrt{2}>3-\sqrt{3}\)