mk làm tắt dc ko

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+...+\frac{5}{90}\)( viết tắt )

\(I=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{9.10}\)

\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(I=5\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(I=5\times\frac{2}{5}\)

\(I=2\)

Vậy \(I=2\)

Tk nha bn ~~

\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)

\(I=\frac{5}{2\cdot3}+\frac{5}{3\cdot4}+\frac{5}{4\cdot5}+\frac{5}{5\cdot6}+\frac{5}{6\cdot7}+\frac{5}{7\cdot8}+\frac{5}{8\cdot9}+\frac{5}{9\cdot10}\)

\(I=5\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)

Theo tính chất của toán HSG lớp 6, ta được

\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(I=5\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(I=5\left(\frac{5}{10}-\frac{1}{10}\right)\)

\(I=5\cdot\frac{4}{10}=5\cdot\frac{2}{5}=\frac{10}{5}=2\)

A=1,319387755 hoặc 1293/980

Đây nha! Tick cho mik nhé! Chỗ mik xuống dòng là viết tiếp đó!

À quên thiếu. Đây mới đúng.

A= 5.(1/2 + 1/6+1/12+1/20+...+1/9506+1/9702+1/9900)

 = 5. (1/1.2 + 1/2.3+1/3.4+1/4.5+...1/97.98+1/98.99+1/99.100)

= 5 .(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/97-1/98+1/98-1/99+1/99-1/100)

= 5.(1-1/100)=5. 99/100=99/20

25

5144 nhe

\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)

\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)

\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)

\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)

\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)

\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)

\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}\)

\(\Rightarrow A=\frac{10}{11}\)

Ta có:

\(A=\frac{3}{2}+\frac{13}{12}+\frac{31}{30}+\frac{57}{56}+\frac{91}{90}\)

\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(=\left(1+1+1+1+1\right)+\left(\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)

\(=5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)

\(B=\frac{5}{6}+\frac{19}{20}+\frac{41}{42}+\frac{71}{72}+\frac{109}{110}\)

\(=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

\(=\left(1+1+1+1+1\right)-\left(\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)

\(=5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\)

=> A - B =\(\left[5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\right]-\left[5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\right]\)

= \(5+\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}-5+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

= \(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(B=\left(1-\frac{1}{6}\right)+\left(1-\frac{19}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

Mk gợi ý đến đây thôi , mk bí rồi đợi mk nghĩ đã!