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Đặt S=1/6+1/12+1/20+1/30+1/42+1/56+1/72
=> S=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> S=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
=> S=1/2-1/9
=> S=7/18
Vì 7/18<1/2
=> S<1/2
Mọi người k mik nhé, :)))
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8-1/9
= 1/2 - 1/9
= 7/18
Bn tự so sánh vs 1/2 nha
A=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A=1/2-1/10
A=2/5
\(S1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(S1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(S1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(S1=1-\frac{1}{10}\)
\(S1=\frac{9}{10}\)
CHÚC BN HC GIỎI !!!!!!!!!! TỨ DIỆP THẢO
S=\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{9.10}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...............+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
1/12+1/6+1/2=(1+2+6)/12=9/12=3/4
1/30+1/20=(3+2)/60=5/6=1/12
1/56+1/42=1/7(1/8+1/6)=1/7(3+4)/24=1/24
8/9-1/72=(8.8-1)/72=63/72=7/8
1/12+1/24=(2+1)/24=3/4
3/4-3/4=0
k cho mik nha!Chúc bn học tốt
\(A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)
\(A=\frac{11}{22}-\frac{2}{22}\)
\(A=\frac{9}{22}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{7}{18}\)
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
=9/10-(1/9*10+1/8*9+...+1/1*2)
=9/10-(1/9-1/10+...+1-1/2)
=9/10-(-1/10+1)=9/10-9/10=0
sua de
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)