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\(=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{41.46}\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{41}-\frac{1}{46}\right)\)
=\(\frac{1}{5}\left(1-\frac{1}{46}\right)=\frac{1}{5}x\frac{45}{46}=\frac{9}{46}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}\)
\(=\frac{2010}{2011}\)
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
= 1 - 1/2011
= 2010/ 2011
Đáp số: 2010/2011
Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
A = \(\frac{24}{48}\)+ \(\frac{12}{48}\)+ \(\frac{8}{48}\)+ \(\frac{2}{48}\)+ \(\frac{1}{48}\)
A = \(\frac{24+12+8+2+1}{48}\)= \(\frac{47}{48}\)
ai tốt bụng thì tk cho mk nha
Ta có \(\left(1-\frac{1}{97}\right)\times\left(1-\frac{1}{98}\right)\times.....\times\left(1-\frac{1}{1000}\right).\)
\(=\frac{97-1}{97}\times\frac{98-1}{98}\times.....\times\frac{1000-1}{1000}\)
\(=\frac{96}{97}\times\frac{97}{98}\times....\times\frac{999}{1000}\) (rút gọn hết )
\(=\frac{96}{1000}\)
\(=\frac{12}{125}\)
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
a,=25/6;7/6=25/6x6/7=25/7
b,=7/2x32/6=56/3
c,=17/5-11/10=34/10-11/10=23/10
d,=8/3+11/4=32/12+33/12=65/12
\(\frac{4}{x}\) \(=\frac{1}{5}\)
Ta có :
\(\frac{4}{x}\) \(=\frac{4}{20}\)
=> \(x=20\)
#Ninh Nguyễn
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)
\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)