Bài 1: Căn bậc hai
Các câu hỏi tương tự
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
1.)\(\sqrt{11+4\sqrt{6}}\)
2.)\(\sqrt{7-4\sqrt{3}}-\sqrt{8+2\sqrt{15}}\)
3.)\(\sqrt{4-2\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
4.)\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
5.)\(\sqrt{4a^2-12a+9}vớia\ge\dfrac{3}{2}\)
6.)\(\sqrt{a^2-6a+9}+\sqrt{9+64a^2-48a}với\dfrac{3}{8}< a< 3\)
Asqrt{19-3sqrt{ }40}-sqrt{19+3sqrt{ }40}
Bsqrt{21-6sqrt{ }6} +sqrt{9+2sqrt{ }18} -2sqrt{6+3sqrt{ }3}
Csqrt{6+2sqrt{ }2sqrt{ }3-sqrt{ }4+2sqrt{ }3}
Dsqrt{4+sqrt{ }15}-sqrt{7-3sqrt{ }5}
Esqrt{2+sqrt{ }3}+sqrt{2-sqrt{ }3}
Fsqrt{12-3sqrt{ }7}-sqrt{12+3sqrt{ }7}
G(3sqrt{2}+sqrt{6}).sqrt{6-3sqrt{ }3}
Hsqrt{9-4sqrt{ }5}-sqrt{14-6sqrt{ }5}
Isqrt{9-4sqrt{ }2}-sqrt{13-4sqrt{ }3}
Đọc tiếp
A=\(\sqrt{19-3\sqrt{ }40}\)-\(\sqrt{19+3\sqrt{ }40}\)
B=\(\sqrt{21-6\sqrt{ }6}\) +\(\sqrt{9+2\sqrt{ }18}\) -2\(\sqrt{6+3\sqrt{ }3}\)
C=\(\sqrt{6+2\sqrt{ }2\sqrt{ }3-\sqrt{ }4+2\sqrt{ }3}\)
D=\(\sqrt{4+\sqrt{ }15}\)-\(\sqrt{7-3\sqrt{ }5}\)
E=\(\sqrt{2+\sqrt{ }3}\)+\(\sqrt{2-\sqrt{ }3}\)
F=\(\sqrt{12-3\sqrt{ }7}\)-\(\sqrt{12+3\sqrt{ }7}\)
G=(3\(\sqrt{2}\)+\(\sqrt{6}\)).\(\sqrt{6-3\sqrt{ }3}\)
H=\(\sqrt{9-4\sqrt{ }5}-\sqrt{14-6\sqrt{ }5}\)
I=\(\sqrt{9-4\sqrt{ }2}\)-\(\sqrt{13-4\sqrt{ }3}\)
Rút gọn các biểu thức :
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
c) \(\sqrt{10+2\sqrt{9}}-\sqrt{9}\)
Rút gọn:
a,\(\frac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\frac{2}{3}\sqrt{12}\)
b,\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\frac{3}{3+\sqrt{6}}\)
c,\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{2}\right)\)
rút gọn biểu thức sau
a/\(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
b/ \(\sqrt{7+3\sqrt{6}}\)+ \(\sqrt{7-2\sqrt{6}}\)- \(2\sqrt{6}\)
c/ \(\sqrt{9+4\sqrt{5}}\) +\(\sqrt{9-4\sqrt{5}}\)- \(2\sqrt{5}\)
Giải phương trình sau:
a, \(\sqrt{9-2\sqrt{14}}+\sqrt{2}\)
b, \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
c,\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
Tính:
a,\(\sqrt{6-2\sqrt{5}}-\sqrt{20}-1\)
b,\(\sqrt{9}+4\sqrt{5}-\sqrt{9-4\sqrt{5}}\)
c,(\(\sqrt{4-2\sqrt{3}}-1).\frac{1}{2\sqrt{3}-4}\)
(cảm ơn)
Rút gọn
a) \(A=\left(\frac{\sqrt{10}-\sqrt{5}}{\sqrt{8}-2}-\frac{\sqrt{90}}{3}\right).\frac{1}{\sqrt{5}}\)
b) \(B=\left(\frac{\sqrt{26}-\sqrt{13}}{1-\sqrt{2}}+\frac{\sqrt{18}-\sqrt{6}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{13}-\sqrt{6}}\)
c) \(C=\frac{\sqrt{10+2\sqrt{21}}-\sqrt{5-2\sqrt{6}}}{\sqrt{9-2\sqrt{14}}}\)