Ta có:
\(B^2=6+\sqrt{11}+6-\sqrt{11}+2\sqrt{\left(6+\sqrt{11}\right)\left(6-\sqrt{11}\right)}=12+2\sqrt{36-11}=12+2\sqrt{25}=12+2.5=22\Rightarrow B=\sqrt{22}\)
Ta có:
\(B^2=6+\sqrt{11}+6-\sqrt{11}+2\sqrt{\left(6+\sqrt{11}\right)\left(6-\sqrt{11}\right)}=12+2\sqrt{36-11}=12+2\sqrt{25}=12+2.5=22\Rightarrow B=\sqrt{22}\)
a) 11+6\(\sqrt{2}\) = \(\left(3+\sqrt{2}\right)^2\)
b) 8-2\(\sqrt{7}\)=\(\left(\sqrt{7}-1\right)^2\)
c)\(\sqrt{11+6\sqrt{2}}=\sqrt{11-6\sqrt{2}}=6\)
d) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
a)\(\sqrt{\left(2\sqrt{2}-3\right)^2+\sqrt{15}}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)}^2+\sqrt{\left(\sqrt{10}-4\right)^2}\)
c)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
d)\(\sqrt{11}+6\sqrt{2}+\sqrt{11-6\sqrt{2}=6}\)
Tính
a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
Thực hiện phép tính (rút gọn biểu thức)
a) \(\sqrt{9+4\sqrt{5}}\) - \(\sqrt{9-4\sqrt{5}}\)
b) \(\sqrt{12-6\sqrt{3}}\) + \(\sqrt{12+6\sqrt{3}}\)
c) \(\sqrt{6\sqrt{2}+11}\) - \(\sqrt{11-6\sqrt{2}}\)
1. Tính
a) \(\sqrt[3]{(\sqrt{2}+3)(11+6\sqrt{2})}\sqrt[3]{(\sqrt{2}+-3)(11-6\sqrt{2})}\)
b) (\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}-\sqrt[3]{2})\)
c)\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
Rút gọn:
a)\(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
b)\(\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
C/m: \(\sqrt{11+6\sqrt{2}}\) + \(\sqrt{11-6\sqrt{2}}\) = 6
\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
Tính \(A=\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}\)