Xét \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Khi đó \(A=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}=1-\frac{1}{\sqrt{2005}}\)
Dạ mọi người giup em bài này với ạ. Dạ em cảm ơn ạ
So sánh
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}\) và 86
Chứng minh : \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{2005\sqrt{2004}}< 2\)
CMR: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+....+\dfrac{1}{2005\sqrt{2004}}< 2\)
a,\(\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}+1}-\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}-1}\)
b,\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
c,\(\frac{2}{\sqrt{3}-\sqrt{5}}+\frac{3-2\sqrt{3}}{\sqrt{3}-2}\)
d,\(\frac{-4}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{3}-1}+\frac{4-2\sqrt{5}}{\sqrt{5}-2}\)
e,\(\frac{6}{\sqrt{5}-1}+\frac{7}{1-\sqrt{3}}-\frac{2}{\sqrt{3}-\sqrt{5}}\)
CMR 2004 < 1+ \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{1006009}}< 2005\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
1,Trục căn thức ở mẫu, rút gọn: ( với \(x\ge0;x\ne1\))
a,\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b,\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
2,Chứng minh các đẳng thức sau:
a,\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=1\)
b,\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
c,\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}+\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=-1\)
d,\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
bài 1:
a) D = \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) E = \(\sqrt[3]{\sqrt{5}-2}+\sqrt[3]{\sqrt{5}+2}\)
c) F =\(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
bài 2:
a) C = \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)
b) D = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\frac{1}{2-\sqrt{3}}\)
c) E =\(\frac{3-x^2}{x+\sqrt{3}}\) với x\(\ne-\sqrt{3}\)
d) F = \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
e) G = \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}\) (có vô hạn dấu căn)
1) Rút gọn:
a) \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
b) \(\frac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}-\frac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
2) Tính A:
A = \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...+\frac{1}{\sqrt{99}-\sqrt{100}}-\frac{1}{\sqrt{100}-\sqrt{101}}\)
Tinh
\(a,\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(b,\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(c,\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(d,\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(e,\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(f,\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(g,\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}\)
