A=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/20)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)+(1-1/90)

A=(1+1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)

A=9-(1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10)

A=9-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

A=9-(1/1-1/10)

A=9-(10/10-1/10)

A=9-9/10

A=90/10-9/10

A=81/10

Tích cho mk nha

đơn giản:

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+.....+\left(1-\frac{1}{90}\right)\)

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

\(A=9-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{5x6}+....+\frac{1}{9x10}\right)\)

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

\(A=9-\left(1-\frac{1}{10}\right)\)

\(A=9-\frac{9}{10}\)

\(A=\frac{90}{10}-\frac{9}{10}\)

\(A=\frac{81}{10}\)

A = 1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90 

A = 1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90 

A = 9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90) 

A = 9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)] 

A = 9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10) 

A = 9 – (1 – 1/10) = 9 – 9/10 = 81/10

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

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

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

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

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

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

\(I=5\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

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

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

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

\(I=2\)

Mấy câu như này tách ra kiểu gì?

\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)

\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{71}{72}+\frac{89}{90}=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=8-\frac{2}{5}=\frac{38}{5}\)

\(A=\left[\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right]:\)\(\frac{1124.2247-1124+1}{1123.2247+1123+1}\)

\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right):\frac{1224.\left(2247-1\right)+1}{1223.\left(2247+1\right)-1}\)

\(A=5^3.\frac{2}{5}:1\)

\(A=5^2.2\)

\(A=50\)

\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)=8+\frac{1}{10}=\frac{81}{10}\)

\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

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

\(=9-\left(1-\frac{1}{10}\right)\)

\(=8+\frac{1}{10}=\frac{81}{10}\)

\(=8+\frac{1}{10}=\frac{81}{10}\)

\(\frac{81}{10}\)

\(=\frac{910}{243}NHA\) Nguyễn Thị Hà My !

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