\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
tính
\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
\(\frac{10}{\sqrt{a}+\sqrt{a+3}}=\frac{10\left(\sqrt{a+3}-\sqrt{a}\right)}{\left(\sqrt{a+3}+\sqrt{a}\right)\left(\sqrt{a+3}-\sqrt{a}\right)}=\frac{10}{3}\left(\sqrt{a+3}-\sqrt{a}\right)\)
tính
\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
tính
\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
=\(\frac{10\left(\sqrt{4}-1\right)}{4-1}+\frac{10\left(\sqrt{7}-1\right)}{7-4}+\frac{10\left(\sqrt{10}-\sqrt{7}\right)}{10-7}+...+\frac{10\left(\sqrt{100}-\sqrt{97}\right)}{100-97}\)
=\(\frac{10}{3}+\frac{10\sqrt{7}-10}{3}+\frac{10\sqrt{10}-10\sqrt{7}}{3}+...+\frac{10\sqrt{100}-10\sqrt{97}}{3}\)
=\(\frac{1}{3}\left(10+10\sqrt{7}-10+10\sqrt{10}-10\sqrt{7}+...+10\sqrt{100}-10\sqrt{97}\right)\)
=\(\frac{1}{3}\left(10\sqrt{100}-10\right)\)
=30
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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Rút gọn biểu thức sau :
A = \(\frac{1}{4\sqrt{1}+1\sqrt{4}}+\frac{1}{7\sqrt{4}+4\sqrt{7}}+\frac{1}{10\sqrt{7}+7\sqrt{10}}...+\frac{1}{2007\sqrt{2004}+2004\sqrt{2007}}\)
tính:
a/\(\frac{6}{4+\sqrt{4-2\sqrt{3}}}\)
b/\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
c/\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{4}+\sqrt{3}}+....+\frac{1}{\sqrt{100}-\sqrt{99}}\)
d/\(\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
\(\frac{6}{\sqrt{7}+2}+\sqrt{\frac{2}{8+3\sqrt{7}}}\)
\(\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
C=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
Trong các dãy số sau, dãy số nào là cấp số cộng? Vì sao?
a) \(10; - 2; - 14; - 26; - 38\)
b) \(\frac{1}{2};\frac{5}{4};2;\frac{{11}}{4};\frac{7}{2}\)
c) \(\sqrt 1 ;\sqrt 2 ;\sqrt 3 ;\sqrt 4 ;\sqrt 5 \)
d) 1; 4; 7; 10; 13
a) Ta có:
\(\begin{array}{l}10 + \left( { - 12} \right) = - 2\\ - 2 + \left( { - 12} \right) = - 14\\ - 14 + \left( { - 12} \right) = - 26\\ - 26 + \left( { - 12} \right) = - 38\end{array}\)
Dãy số là cấp số cộng
b) Ta có:
\(\begin{array}{l}\frac{1}{2} + \frac{3}{4} = \frac{5}{4}\\\frac{5}{4} + \frac{3}{4} = 2\\2 + \frac{3}{4} = \frac{{11}}{4}\\\frac{{11}}{4} + \frac{3}{4} = \frac{7}{2}\end{array}\)
Dãy số là cấp số cộng
c) Không xác định được d giữa các số hạng
Dãy số không là cấp số cộng
d) Ta có:
\(\begin{array}{l}1 + 3 = 4\\4 + 3 = 7\\7 + 3 = 10\\10 + 3 = 13\end{array}\)
Dãy số là cấp số cộng
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)