\(\sqrt[3]{37-30\sqrt{3}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
Ta có: \(E=\sqrt{37-6\sqrt{30}}=\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}=\left|3\sqrt{3}-\sqrt{10}\right|=3\sqrt{3}-\sqrt{10}\)
\(F=\sqrt{51-6\sqrt{30}}=\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}=\left|3\sqrt{5}-\sqrt{6}\right|=3\sqrt{5}-\sqrt{6}\)
\(G=\sqrt{59-6\sqrt{30}}=\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}=\left|3\sqrt{6}-\sqrt{5}\right|=3\sqrt{6}-\sqrt{5}\)
\(H=\sqrt{17-2\sqrt{30}}=\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}=\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)
\(E=\sqrt{37-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}\\ =\left|3\sqrt{3}-\sqrt{10}\right|\\ =3\sqrt{3}-\sqrt{10}\)
\(F=\sqrt{51-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}\\ =\left|3\sqrt{5}-\sqrt{6}\right|\\ =3\sqrt{5}-\sqrt{6}\)
\(G=\sqrt{59-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}\\ =\left|3\sqrt{6}-\sqrt{5}\right|\\ =3\sqrt{6}-\sqrt{5}\)
\(H=\sqrt{17-2\sqrt{30}}\\ =\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}\\ =\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)
\(\sqrt{37-20\sqrt{3}}+\sqrt{37+20\sqrt{3}}\)
\(=\sqrt{37-2\sqrt{300}}+\sqrt{37+2\sqrt{300}}\)
\(=\sqrt{\left(5-\sqrt{12}\right)^2}+\sqrt{\left(5-\sqrt{12}\right)^2}\)
\(=|5-\sqrt{12}|+|5+\sqrt{12}|\)
\(=5-\sqrt{12}+5+\sqrt{12}\)
\(=10\)
= \(\sqrt{12-2.2\sqrt{3}.5+25}-\sqrt{12+2.2\sqrt{3}.5+25}\)
= \(\sqrt{\left(2\sqrt{3}-5\right)^2}-\sqrt{\left(2\sqrt{3}+5\right)^2}\)
= \(|2\sqrt{3}-5|-2\sqrt{3}-5\)
=\(5-2\sqrt{3}-2\sqrt{3}-5=-4\sqrt{3}\)
\(=\sqrt{\left(5-2\sqrt{3}\right)^2}+\sqrt{\left(5+2\sqrt{3}\right)^2}\)
\(=\left|5-2\sqrt{3}\right|+\left|5+2\sqrt{3}\right|\)
\(=5-2\sqrt{3}+5+2\sqrt{3}\)
\(=10\)
\(2\sqrt[]{37+20\sqrt[]{3}}-\sqrt[]{73-40\sqrt[]{3}}\)
\(=2\sqrt[]{25+2.5.2\sqrt[]{3}+12}-\sqrt[]{48-2.5.4\sqrt[]{3}+25}\)
\(=2\sqrt[]{\left(5+2\sqrt[]{3}\right)^2}-\sqrt[]{\left(5-4\sqrt[]{3}\right)^2}\)
\(=2\left|5+2\sqrt[]{3}\right|-\left|5-4\sqrt[]{3}\right|\)
\(=2\left(5+2\sqrt[]{3}\right)-\left(4\sqrt[]{3}-5\right)\left(vì.4\sqrt[]{3}>5\right)\)
\(=10+4\sqrt[]{3}-4\sqrt[]{3}+5\)
\(=15\)