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Giải:
\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}\)
\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}\)
\(=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|\)
\(=5-\sqrt{17}+\sqrt{17}-4\)
\(=1\)
Vậy ...
\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}=\sqrt{25-2.5.\sqrt{17}+17}+\sqrt{16-2.4.\sqrt{17}+17}=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|=5-\sqrt{17}+\sqrt{17}-4=1\)
\(\sqrt{17-\sqrt{33}}\cdot\sqrt{17+\sqrt{33}}\)
\(=\sqrt{\left(17-\sqrt{33}\right)\left(17+\sqrt{33}\right)}\)
\(=\sqrt{17^2-\left(\sqrt{33}\right)^2}\)
\(=\sqrt{289-33}\)
\(=\sqrt{256}\)
\(=\sqrt{16^2}\)
\(=16\)
\(\sqrt{17-\sqrt{33}}\cdot\sqrt{17+\sqrt{33}}\)
\(=\sqrt{17^2-\left(\sqrt{33}\right)^2}\)
\(=\sqrt{289-33}=\sqrt{256}=16\)
Có :
+) \(\sqrt{33}< \sqrt{36}\)
+) \(\sqrt{17}>\sqrt{15}\Rightarrow-\sqrt{17}< -\sqrt{15}\)
Cộng theo vế 2 bất pt :
\(\sqrt{33}-\sqrt{17}< \sqrt{36}-\sqrt{15}=6-\sqrt{15}\)
Vậy...
Có :
\(3\sqrt{2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{12}\)
Mà \(\sqrt{18}>\sqrt{12}\Rightarrow3\sqrt{2}>2\sqrt{3}\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Lời giải:
Xét hiệu:
\(\sqrt{33}-\sqrt{17}-(6-\sqrt{15})=(\sqrt{33}-6)+(\sqrt{15}-\sqrt{17})\)
\(< (\sqrt{36}-6)+(\sqrt{17}-\sqrt{17})=0+0=0\)
\(\Rightarrow \sqrt{33}-\sqrt{17}< 6-\sqrt{15}\)
------------------------
\(\sqrt{3\sqrt{2}}=\sqrt{\sqrt{3^2.2}}=\sqrt[4]{18}\)
\(\sqrt{2\sqrt{3}}=\sqrt{\sqrt{2^2.3}}=\sqrt[4]{12}\)
Mà \(18>12\Rightarrow \sqrt[4]{18}>\sqrt[4]{12}\Rightarrow \sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
\(\sqrt{37}-\sqrt{15};2\)
có \(\left(\sqrt{37}-\sqrt{15}\right)^2=37-2\sqrt{555}+15=52-2\sqrt{555}\)
\(2^2=4\)
xét \(52-2\sqrt{555}-4=48-2\sqrt{555}\)
SS:\(48;2\sqrt{555}\)
\(48^2=2304\)
\(\left(2\sqrt{555}\right)^2=2220\)
2304>2220=>\(\sqrt{37}-\sqrt{15}>2\)
\(\sqrt{17-\sqrt{33}}\sqrt{17+\sqrt{33}}=\sqrt{\left(17-\sqrt{33}\right)\left(17+\sqrt{33}\right)}\)
\(=\sqrt{17^2-33}=\sqrt{256}=16\)