Ta có
\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\) và \(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\). Tính A : B
làm giúp mình bài này nhék
\(A=\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-.....-\frac{1}{2.3}-\frac{1}{1.2}\)
Mình cảm ơn nhiều mình sẽ ****
CMR: \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\). CMR: \(\frac{7}{12}< A< \frac{5}{6}\)
Tính:\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
CMR hang dang thuc thuc nao sau voi n la so tu nhien :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\left(n\ge2\right)\)
Tính
( 1 - \(\frac{2}{2.3}\) ) . ( 1 - \(\frac{2}{3.4}\) ) . ( 1 - \(\frac{2}{4.5}\) ) . . . . . ( 1 - \(\frac{2}{99.100}\) )
\(\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...\frac{1}{200}}=1\)
Hãy chứng minh
Tính giá trị của biểu thức:
a) A= (153 + 5. 152 - 53) : ( 183 + 6. 182 - 63)
b) \(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\)
c) \(C=\frac{-1}{1.2}.\frac{-2^2}{2.3}.\frac{-3^2}{3.4}...\frac{-99^2}{99.100}.\frac{-100^2}{100.101}\)
