sửa đề: \(\sqrt{6+\sqrt{35}}.\sqrt{6-\sqrt{35}}=1\)
giải:
\(\sqrt{6+\sqrt{35}}.\sqrt{6-\sqrt{35}}=\sqrt{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}\\ \sqrt{36-35}=\sqrt{1}=1\left(đpcm\right)\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3\left(\sqrt{6}-4\right)}}\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3}\left(\sqrt{6}-4\right)}\)
\(B=2+3\sqrt{2}-2\sqrt{32}-\sqrt{6+4\sqrt{2}}\)
\(G=\sqrt{12-2\sqrt{35}}+4\sqrt{20}+\sqrt{28}\)
\(E=\left(\frac{2\sqrt{2}+3\sqrt{3}}{\sqrt{2}+\sqrt{3}}-\sqrt{6}\right):\left(\sqrt{2}-\sqrt{3}\right)-\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
Chứng minh rằng:
\(B=\sqrt{5\sqrt{5}\sqrt{5}...\sqrt{5}\sqrt{5}}+\sqrt{6+\sqrt{6}+\sqrt{6}+...\sqrt{6}+\sqrt{6}}< 8\)
Chứng minh: \(\sqrt{5\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6...+\sqrt{6}}}}}}\le8\) có 2018 dấu căn ở mỗi hạng tử
Chứng minh \(4< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}< 5}\)
Chứng minh: \(4< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 5\)
Bài 1: Rút gọn các biểu thức
a)\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
b)\(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
c)\(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
d)\(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
e)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\left(\sqrt{x}-\sqrt{y}\right)^2\)
Chứng minh: (\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)=-14,5\sqrt{2}\)
