Question

[❃๖ۣۜY๖ۣۜi๖ۣۜn ⓛ ⓞ ⓥ ⓔ ♡]

So sánh \(x=\sqrt{3}+\sqrt{6}\) và \(y=\sqrt{2}+\sqrt{7}\)

Answer 1

[soyeon_Tiểubàng giải]

Ta so sánh: \(\sqrt{3}-\sqrt{2}\) và \(\sqrt{7}-\sqrt{6}\)

\(\sqrt{3}-\sqrt{2}=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\frac{3-2}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}\)

\(\sqrt{7}-\sqrt{6}=\frac{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{\sqrt{7}+\sqrt{6}}=\frac{7-6}{\sqrt{7}+\sqrt{6}}=\frac{1}{\sqrt{7}+\sqrt{6}}\)

Vì \(\sqrt{3}+\sqrt{2}< \sqrt{7}+\sqrt{6}\)

nên \(\frac{1}{\sqrt{3}+\sqrt{2}}>\frac{1}{\sqrt{7}+\sqrt{6}}\)

\(\Rightarrow\sqrt{3}-\sqrt{2}>\sqrt{7}-\sqrt{6}\)

\(\Rightarrow\sqrt{3}+\sqrt{6}>\sqrt{7}+\sqrt{2}\) hay x > y