\(\sqrt{10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}\)
\(=\sqrt{5+3+2+2\sqrt{5}.\sqrt{3}+2\sqrt{3}.\sqrt{2}+2\sqrt{5}.\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
\(\sqrt{10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}\)
\(=\sqrt{5+3+2+2\sqrt{5}.\sqrt{3}+2\sqrt{3}.\sqrt{2}+2\sqrt{5}.\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
Giải các pt
a, \(\sqrt{x-1}=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}-\sqrt{2\sqrt{6}+5}\)
b,\(\sqrt{2x-2\sqrt{x^2-4}}+\sqrt{x-2}=4\)
❤ Tính:
a) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
b)\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
c)\(\sqrt{7+\sqrt{24}}+\sqrt{31-\sqrt{600}}\)
d)\(\sqrt{28-\sqrt{300}}+\sqrt{4-\sqrt{12}}\)
e)\(\sqrt{7-\sqrt{40}}-\sqrt{5-\sqrt{24}}-\sqrt{6-\sqrt{20}}\)
f)\(\sqrt{48-10\sqrt{7+\sqrt{48}}}\)
g) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{15}-\sqrt{16}}\)
Chứng minh rằng
\(5\sqrt{2}< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{50}}< 10\sqrt{2}\)
Chứng minh rằng:
\(5\sqrt{2} < 1+\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} +...+\dfrac{1}{\sqrt{50}} < 10\sqrt{2}\)
Cho \(A=\sqrt{20+\sqrt{20+\sqrt{20+.....+\sqrt{20}}}}\)
\(B=\sqrt[3]{24+\sqrt[3]{24+....+\sqrt[3]{24}}}\)
Chứng minh rằng 7 < A + B < 8
Bài 1: Rút gọn biểu thức
a) \(A=\sqrt{26+15\sqrt{3}}\)
b) \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
c) \(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
d) \(D=\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
e) \(E=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\)
f) \(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
g) \(G=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
h) \(H=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
Chứng minh các đẳng thức:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)=1
b)\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)-1 =0
c) \(\sqrt{26+15\sqrt{3}}+\sqrt{26-15\sqrt{3}}-5\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\)
Rút gọn các biểu thức sau:
1) \(\frac{1}{\sqrt{7-\sqrt{24}+1}}-\frac{1}{\sqrt{7+\sqrt{24}}}\)
2) \(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
3) \(\sqrt{\frac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+\sqrt{6}}}\)
4) \(\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\)
Cho A=\(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{2\left(\sqrt{3}+\sqrt{5}\right)}+\frac{1}{3\left(\sqrt{5}+\sqrt{7}\right)}+...+\frac{1}{40\left(\sqrt{79}+\sqrt{81}\right)}\)
Chứng minh rằng A<\(\frac{8}{9}\)
Giúp mình với, mình đang rối quá