Chứng tỏ rằng \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=1.3.5...59\)
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TG
1
3 tháng 5 2015
Ta có:
\(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2^{30}}\)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}\)
\(=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{\left(1.3.5.7...59\right)\left(2.4.6.8...60\right)}{\left(2.4.6.8...60\right)}\)
\(=1.3.5.7...59\)
Vậy \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=1.3.5.7...59\)
NH
0
9 tháng 7 2016
\(60!=1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot59\cdot60=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2\cdot4\cdot6\cdot...\cdot58\cdot60\)
\(=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\cdot1\cdot2\cdot3\cdot...\cdot30=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\times30!\)
\(\Rightarrow1\cdot3\cdot5\cdot...\cdot59=\frac{60!}{30!\times2^{30}}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}\)đpcm.
9 tháng 7 2016
\(\frac{31}{2}\cdot\frac{32}{2}\cdot...\cdot\frac{60}{2}\cdot2\cdot4\cdot...\cdot58\cdot60\)
=31.32.33.34...60.1.2.3.4.5...29.30
=1.2.3.4.5.6.7.8.9.10...60
1.3.5.7...59.2.4.6.8...60
=1.2.3.4.5.6...60
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}=1\cdot3\cdot5\cdot...\cdot59\)
LT
3
11 tháng 5 2015
\(1.3.5....59=\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}=\frac{\left(1.2.3.4.5...30\right).31....59.60}{2^{30}.\left(1.2.3...30\right)}=\frac{31.32....60}{2^{30}}=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}\)
LH
0
BV
1
22 tháng 6 2018
Ta có: \(\frac{31}{2}.\frac{32}{2}...\frac{60}{2}=\frac{31.32...60}{2^{30}}=31.33...57.59.\left(\frac{32.34...58.60}{2^{30}}\right)\)
\(=31.33...57.59.\left(\frac{16.17...29.30}{2^{15}}\right)=17.19...27.29.31.33...57.59.\left(\frac{16.18...30}{2^{15}}\right)\)
\(=17.19...57.59.\left(\frac{8.9...15}{2^7}\right)=9.11.13.15.17...57.59.\left(\frac{8.10.12.14}{2^7}\right)\)
\(=9.11...57.59.\left(\frac{4.5.6.7}{2^3}\right)=5.7.9...57.59.\left(\frac{4.6}{2^3}\right)\)
\(=5.7.9...57.59.3=1.3.5...59\)
NT
1
26 tháng 5 2015
\(1.3.5.7.9...59=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{2.4.6...60}=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{31.32.33...60}{2.2.2...2}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}\)
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}=1.3.5...59\)(đpcm)
PL
3
25 tháng 9 2017
\(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)
\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)
\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)
Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)
25 tháng 9 2017
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
NP
0
Ta có: \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2.2.2...2}=\frac{31.32.33...60}{2^{30}}\)
(30 số 2)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}=\frac{31.32.33...60.1.2.3...30}{\left(2.1\right).\left(2.2\right).\left(2.3\right)...\left(2.30\right)}=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{\left(2.4.6...60\right)}=1.3.5...59\)
\(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2^{30}}\)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}\)
\(=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{\left(1.3.5.7...59\right)\left(2.4.6.8...60\right)}{\left(2.4.6.8...60\right)}\)
\(=1.3.5.7...59\)