chứng tỏ rằng:
31/2*32/2*33/2*...*60/2=1.3.5.7....59
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DD
Đoàn Đức Hà
Giáo viên
8 tháng 10 2021
\(1.3.5.7.....59=\frac{1.2.3.4.....60}{2.4.6.....60}=\frac{\left(1.2.3.....30\right).\left(31.32.....60\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(30.2\right)}\)
\(=\frac{\left(1.2.3.....30\right).\left(31.32.....60\right)}{\left(1.2.3.....30\right).\left(2.2.....2\right)}=\frac{31}{2}.\frac{32}{2}.....\frac{60}{2}\)
NP
0
NB
0
YC
2
3 tháng 5 2015
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)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}=\frac{31.32.33...60.1.2.3...30}{\left(2.1\right).\left(2.2\right).\left(2.3\right)...\left(2.30\right)}=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{\left(2.4.6...60\right)}=1.3.5...59\)
3 tháng 5 2015
\(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2^{30}}\)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}\)
\(=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{\left(1.3.5.7...59\right)\left(2.4.6.8...60\right)}{\left(2.4.6.8...60\right)}\)
\(=1.3.5.7...59\)
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
11 tháng 6 2015
Ta có:
31/2.32/2.33/2....60/2=31.32......60/2^30
=(31.32.33....60)(1.2.3....30)/2^30(1.2.3...30)
=(1.3.5...59)(2.4.6...60)/(2.4.6...60)=1.3.5...59
=>P=Q
nhớ ****
VN
6 tháng 4 2017
cái dòng 3, 4 mk ko hiểu sao 2^30.(1.2.3....30) lại bằng 2.4.6...60
LM
13 tháng 6 2018
Ta có:\(\dfrac{31}{2}\).\(\dfrac{32}{2}\).\(\dfrac{33}{2}\).....\(\dfrac{60}{2}\)
=\(\dfrac{31.32.33.....60}{2^{30}}\)
=\(\dfrac{\left(1.2.3.....30\right).\left(31.32.33.....60\right)}{\left(1.2.3.....30\right).2^{30}}\)
=\(\dfrac{1.2.3.....60}{2.4.6.....60}\)
=\(\dfrac{\left(1.3.5.....59\right).\left(2.4.6.....60\right)}{2.4.6.....60}\)
=1.3.5.....59
Vậy (đpcm)
Ta có : 1.3.5.7. ... .59
= \(\frac{1.2.3.4.5.....58.59.60}{2.4.6.8.....58.60}\)
= \(\frac{\left(1.2.3.....30\right).\left(31.32.....60\right)}{\left(1.2.3.....30\right).2.2.....2}\)
= \(\frac{31.32.....60}{2.2.....2}\)
= \(\frac{31}{2}.\frac{32}{2}.....\frac{58}{2}.\frac{60}{2}\) ( đpcm )