Question

so sánh

a) căn 6 - căn 7 và căn 7 - căn 8

Answer 1

ta se chứng minh

\(\sqrt{6}+\sqrt{8}< 2\sqrt{7}.Mà:\sqrt{6}+\sqrt{8};2\sqrt{7}đêu>0\Rightarrow\sqrt{6}+\sqrt{8}< 2\sqrt{7}\Leftrightarrow\left(\sqrt{6}+\sqrt{8}\right)^2< \left(2\sqrt{7}\right)^2\Leftrightarrow14+2\sqrt{48}< 28;14+2\sqrt{48}< 14+2\sqrt{49}=14+14=28\Rightarrow\sqrt{6}+\sqrt{8}< 2\sqrt{7}\left(1\right)\)\(\sqrt{6}-\sqrt{7}-\left(\sqrt{7}-\sqrt{8}\right)=\sqrt{6}+\sqrt{8}-2\sqrt{7}=\left(\sqrt{6}+\sqrt{8}\right)-2\sqrt{7}< 0\left(\sqrt{6}+\sqrt{8}< 2\sqrt{7}theo\left(1\right)\right)\) \(\Rightarrow\sqrt{6}-\sqrt{7}< \sqrt{7}-\sqrt{8}\)