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\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(\Rightarrow S=1-\frac{1}{46}\)
\(\Rightarrow S=\frac{45}{46}\)
Bài làm
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(S=\frac{1}{1}-\frac{1}{46}\)
\(S=\frac{46}{46}-\frac{1}{46}\)
\(S=\frac{45}{46}\)
Vậy \(S=\frac{45}{46}\)
# Học tốt #
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)\)
\(A=1-\frac{1}{9}=\frac{8}{9}\)
A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
=1\(-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}=\frac{8}{9}\)
Vậy A=\(\frac{8}{9}\)
chứng tỏ :
Ta có : \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
áp dụng :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=1-\frac{1}{9}\)
\(A=\frac{8}{9}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=1-\frac{1}{9}=\frac{8}{9}\)
Cho A = 1 + 2 + 22 + 23 + ... + 22008
-> 2A = 2 + 22 + 23 + 24 +...+ 22009
-> 2A - A = ( 2 + 22 + 23 + 24 +...+ 22009 ) - ( 1 + 2 + 22 + 23 + ... + 22008 )
-> A = \(2^{2009}-1=-\left(1-2^{2009}\right)\)
S = \(\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}\)=-1
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
=1-1/9
=8/9
A) \(x = {7 \over 10}- {8 \over10} \)
\(x = {-1 \over 10}\)
B)\({2 \over3}x = 2{5 \over 6}-{3 \over4}\)
\({2 \over3}x = {25 \over 12}\)
\(x = {25 \over 12}/{2 \over3} \)
\(x = {25\over 8}\)
2/ Tính tổng:
\( = {8 \over 9}\)
Ta có:
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{870}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{29.30}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{2}-\frac{1}{30}=\frac{15}{30}-\frac{1}{30}=\frac{14}{30}=\frac{7}{15}\)
Vậy \(A=\frac{7}{15}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{870}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{29.30}\)
\(A=\frac{1}{2}-\frac{1}{30}\)
\(A=\frac{7}{15}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(A=1-\frac{1}{21}\)
\(A=\frac{20}{21}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\)
\(A=1-\frac{1}{21}\)
\(A=\frac{20}{21}\)