Question

so sánh

Answer 1

\(A=3-2\sqrt{2}=\dfrac{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{3+2\sqrt{2}}=\dfrac{9-8}{3+2\sqrt{2}}=\dfrac{1}{\sqrt{9}+\sqrt{8}}\)

\(B=2\sqrt{2}-\sqrt{7}=\dfrac{\left(2\sqrt{2}-\sqrt{7}\right)\left(2\sqrt{2}+\sqrt{7}\right)}{2\sqrt{2}+\sqrt{7}}=\dfrac{8-7}{2\sqrt{2}+\sqrt{7}}=\dfrac{1}{\sqrt{7}+\sqrt{8}}\)

Do \(\sqrt{9}+\sqrt{8}>\sqrt{7}+\sqrt{8}\) nên \(\dfrac{1}{\sqrt{9}+\sqrt{8}}< \dfrac{1}{\sqrt{7}+\sqrt{8}}\)

\(\Rightarrow A< B\)