Ta có:
\(\sqrt{6}+\sqrt[3]{6}< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< \sqrt{6+\sqrt{6+...+\sqrt{9}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{8}}}\)
\(\Leftrightarrow4< \sqrt{6}+\sqrt[3]{6}< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 3+2\)
\(\Leftrightarrow4< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 5\)
CMR:\(4< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 5\)
Chứng minh 4<\(\sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 5\)
Bài 1:
1.\(\sqrt{2-\sqrt{3}}\)
2.\(\sqrt{3+\sqrt{5}}\)
3.\(\sqrt{21-6\sqrt{6}}\)
4.\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
5.\(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}\)
6.\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
Tính:
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)
c) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
Gỉai giúp mk vs
\(\sqrt{21-6\sqrt{6}}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}\)
\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(\sqrt{6}\left(\sqrt{26+15\sqrt{3}}+\sqrt{26-15\sqrt{3}}\right)\)
a)\(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\frac{6\sqrt{2}-4}{3-\sqrt{2}}\)
b)\(\sqrt{2-\sqrt{3}}-\sqrt{\frac{3}{2}}\)
c)\(\frac{\sqrt{30}-\sqrt{2}}{\sqrt{8-\sqrt{15}}}-\sqrt{8-\sqrt{49+8\sqrt{3}}}\)
d) \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
e)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
f)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
g)\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
thực hiện phép tính
A=\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2-\sqrt{2-\sqrt{3}}}}\)
B=\(\dfrac{6+4\sqrt{2}}{\sqrt{2+\sqrt{6+4\sqrt{2}}}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
rút gọn
\(\frac{\sqrt{3}\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
b, \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(\left(\dfrac{15}{3-\sqrt{3}}-\dfrac{2}{1-\sqrt{3}}+\dfrac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
