Chứng tỏ 1.3.5.7.9....197.199=101/2.102/2.103/2....200/2
Chứng tỏ :1.3.5.7.9....197.199=101/2.102/2.103/2.....200/2
chứng tỏ :1.3.5.7.....197.199=101/2.102/2.103/2....200/2
\(\dfrac{101}{2}.\dfrac{102}{2}.\dfrac{103}{2}.\dfrac{104}{2}.....\dfrac{200}{2}\\ =\dfrac{101.102.103.104.....200}{2^{100}}\\ =\dfrac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}.\left(1.2.3.....100\right)}\\ =\dfrac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right).....\left(2.100\right)}\\ =\dfrac{\left(1.3.5.....199\right)\left(2.4.6.....200\right)}{4.6.8.....200}\\ =1.3.5.7.....197.199\)
=> Điều phải chứng minh
chứng tỏ rằng:1.3.5.7...197.199=101/2.102/2.103/2...200/2
\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)
\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)
\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)
\(=1.3.5.....199\)
Chứng tỏ rằng: 1.3.5.7.9. ... .197.199 = 101/2 . 102/2 . 103/2 . ... . 200/2
Giúp mình nhé các bạn!
Ta có :
\(1.3.5.7.....199\)
\(=\frac{1.2.3.4.5.6.7.....198.199.200}{2.4.6.....198.200}\)
\(=\frac{\left(1.2.3.....99.100\right)\left(101.102.....200\right)}{\left(1.2.3.....99.100\right)\left(2.2.2.....2.2\right)}\)
\(=\frac{101.102.....200}{2.2.....2}\)
\(=\frac{101}{2}.\frac{102}{2}.....\frac{200}{2}\left(đpcm\right)\)
Chứng minh rằng:
1.3.5.7.9.....197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+...+\frac{200}{2}\)
Chứng tỏ rằng :1.5.7...197.199=\(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}...\frac{200}{2}\)
#)Giải :
Ta có : \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}=\frac{101.102.103.....200}{2^{100}}=\frac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}\left(1.2.3.....100\right)}\)
\(=\frac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.100\right)}=\frac{\left(1.3.5.....99\right)\left(2.4.6.....100\right)}{2.4.6.....200}=1.3.5.....99\left(đpcm\right)\)
Ta có : 1.3.5.7.....199 = \(\frac{\left(1.3.5.7.....199\right).\left(2.4.6.8.....200\right)}{2.4.6.8.....200}=\frac{1.2.3.4.5.....199.200}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}=\frac{1.2.3.4.5.....199.200}{2^{100}.1.2.3.....100}=\frac{101.102.103.....200}{2^{100}}\)\(=\frac{101}{2}.\frac{102}{2}\frac{103}{2}.....\frac{200}{2}\)\( \left(ĐPCM\right)\)
chứng minh rằng : 1.3.5.7....197.199 = \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}....\frac{200}{2}\)
1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)
CMR: 1.3.5.7......197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+........+\frac{200}{2}\)
CHứng tỏ:
1.3.5.7.9...97.99 = 51/2 . 52.2 . 53/2 ... 99/2 . 100/2
đặt A = 1.3.5.7.9...97.99,B=51/2 . 52/2 . 53/2 ... 99/2 . 100/2
A=1.3.5.7.9....97.99
=1.3.5.7.....97.99.2.4.6...100/2.4.6...100
=1.2.3.4.5...100/2.1.2.2.2.3.2.4...2.50
=1.2.3.4.5...100/1.2.3.4...50.2.2.2...2(50 chữ số 2)
=51.52.53...100/2.2.2.2...100
B=51/2.52/2.53/2...99/2.100/2
Suy ra A=B
Bạn nhân cả 2 vế với 1 lượng 2 . 4 . 6 ..... 98 . 100