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\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(S=1-\frac{1}{9}=\frac{8}{9}\)

a) \(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{23}{60}\)
b) \(\frac{5}{2}+\frac{1}{3}-\frac{1}{4}\)
\(=\frac{31}{12}\)
c) \(\frac{10}{3}-\left(\frac{1}{4}+\frac{1}{12}\right)\)
\(=\frac{9}{3}=3\)
d) \(\frac{1}{3}+\frac{6}{5}-\frac{1}{15}\)
\(=\frac{22}{15}\)
trả lòi:
A)= 23/60
B)= 31/12
C)= 3* vì 3/9 cùng chia hết cho 3*
D)= 22/15
Hc tốt!

1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9=135


a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)

\(=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Ta có: \(\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19\)
\(=\left(\frac12+\frac13+\frac16\right)+\left(\frac14+\frac18\right)+\left(\frac15+\frac17+\frac19\right)\)
\(=1+\frac28+\frac18+\left(\frac{63}{315}+\frac{45}{315}+\frac{35}{315}\right)\)
\(=1+\frac38+\frac{143}{315}=\frac{11}{8}+\frac{143}{315}=\frac{11\cdot315+143\cdot8}{8\cdot315}=\frac{4609}{2520}\)