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ta lấy ví đụ 1/2
vì 1/2 đã nhỏ hơn 1 mà các số kia đều nhỏ hơn 1/2
k nhé
đoạn cuối cùng là lớn hơn 1 chứ ko phải 11 nhe mình đánh nhầm . xin lỗi
Thấy 1/41+1/42 +......+ 1/60 < 1/40 .20
1/41 +1/42 + .....+1/60<1/2
mà 1/61 +1/62+......+1/80 < 1/60 .20 =1/3
suy ra 1/41+1/42+ .......+1/80 <1/2 +1/3=7/12(đpcm)
Lại có 1/41 +1/42 +.....+1/80 <1/40 .40 =1(đpcm)
\(=1-\frac{1}{1\cdot2}+1-\frac{1}{2\cdot3}+1-\frac{1}{3\cdot4}+...+1-\frac{1}{9\cdot10}\)
\(=\left[1+1+1+...+1\right]-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right]\)
\(=9-\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right]=9-\left[1-\frac{1}{10}\right]\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}\)
\(=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)\)
\(=\left(1+1+1+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=4-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=4-\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(=4-\left(\frac{1}{2}-\frac{1}{7}\right)\)
\(=4-\frac{5}{14}\)
\(=\frac{51}{14}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}\)
\(=>\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\frac{350}{420}+\frac{385}{420}+\frac{399}{420}+\frac{406}{420}+\frac{410}{420}\)
\(=>\frac{350}{420}+\frac{385}{420}+\frac{399}{420}+\frac{406}{420}+\frac{410}{420}=\frac{350+385+399+406+410}{420}\)
\(=>\frac{350+385+399+406+410}{420}=\frac{1950}{420}=\frac{65}{14}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{!}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=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
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=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)
=9 – (1 – 1/10) = 9 – 9/10
= 81/10
\(y=\frac{1+5+11+19+29+41+55+71+89}{2+6+12+20+30+42+56+72+90}\)
\(y=\frac{1x2-1+2x3-1+3x4-1+4x5-1+5x6-1+6x7-1+7x8-1+8x9-1+9x10-1}{1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10}\)
\(y=\frac{\left(1x2+2x3+...+9x10\right)-\left(1+1+1+1+1+1+1+1+1\right)}{1x2+2x3+...+9x10}\)
\(y=\frac{1x2+2x3+...+9x10}{1x2+2x3+...+9x10}-\frac{9}{1x2+2x3+...+9x10}\)
\(y=1-\frac{9}{1x2+2x3+...+9x10}\)
Chọn D