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HP
10 tháng 1 2016
31/2.32/2.33/2....60/2=(31.32.33...60)/230
=[(31.32.33...60).(1.2.3...30)]/230.(1.2.3...30)
=[(1.3.5...59).(2.4.6....60)]/(2.4.6...60)=1.3.5...59
30 tháng 10 2016
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{49}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10< S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50.10}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{3}{20}< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{4}{15}+\frac{1}{5}\)
\(\frac{3}{5}< S< \frac{4}{5}\left(đpcm\right)\)
D
0
DP
21 tháng 7 2017
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{32}\right)+\left(1+\frac{1}{64}\right)-7\)
\(S=\left(1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-7\)
\(S=6+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+....+\left(\frac{1}{32}-\frac{1}{64}\right)\right]-7\)
\(S=6+\left(1-\frac{1}{64}\right)-7\)
\(S=6+\frac{63}{64}-7\)
\(S=\frac{447}{64}-7=-\frac{1}{64}\)
SN
15 tháng 11 2016
chịu thui
chúc bn học gioi!
nhaE@@
Toán lớp 7 bye mk đi hc đây hihi
$$$$
21 tháng 3 2017
Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
= \(\frac{1}{3}+\)( \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\)) \(+\)( \(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\) ) \(< \)\(\frac{1}{3}+\)( \(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)) \(+\)( \(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\)) = \(\frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)
\(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)
\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)
\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)
Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)
ta có:Đặt A= \(1.3.5.....59=\frac{1.2.3.4.....59.60}{2.4.6.....60}\)
=\(\frac{1.2.3.....59.60}{2^{30}.\left(1.2.3.....30\right)}=\frac{31.32.....59.60}{2^{30}}\)
= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
vì \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\) = \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
\(\Rightarrow\)A= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
( Điều phải chứng minh)
toán nâng cao lớp 6 đấy bạn nha