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\(S=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{90}+\frac{1}{100}\)
\(S=\left(\frac{1}{6}+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{100}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)\)
\(S=\left(\frac{2}{12}+\frac{1}{12}\right)+\left(\frac{5}{100}+\frac{1}{100}\right)+\left(\frac{3}{90}+\frac{1}{90}\right)\)
\(S=\frac{3}{12}+\frac{6}{100}+\frac{4}{90}\)
\(S=\frac{1}{4}+\frac{3}{50}+\frac{2}{45}\)
\(S=\frac{319}{900}\)
S=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.15}+..\)
\(S=SAI\)
HÌNH NHƯ SAI RỒI
HÌNH NHỨIA

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chú ý : Dấu '' . '' là dấu nhân
CHÚC BN HỌC TỐT !!!! ^_^

Gọi tổng dãy số hạng trên là A
A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)
Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số
A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\)
A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)

\(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{90}\)
\(=5.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=5.\left(1-\frac{1}{10}\right)\)\
\(=5.\frac{9}{10}\)
\(=\frac{9}{2}\)

Ta có : \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=\frac{81}{10}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)

Chỗ cuối mk nhầm, sửa lại nha :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}+\frac{1}{100}\)
\(=\frac{41}{100}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}-\frac{1}{100}\)
\(=\frac{39}{100}\)
Đúng thì k nha bn !!!!

Đặt biểu thức đó là A A = 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/90 A = 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + ... + 1/9x10 A = 1/2 - 1/3 + 1/3 - 1/4 + 14 - 1/5 + 1/5 - 1/6 + ..+ 1/9 - 1/10 A = 1/2 - 1/10 A = 2/5
1/6 + 1/12 + 1/20 + 1/30 + ... + 1/90
= 1/2×3 + 1/3×4 +1/4×5 + 1/5×6 + ... × 1/9×10
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/9 - 1/10
= 1/2 - 1/10
= 2/5
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