\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)

\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)

\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)

\(=1.3.5.....199\)

sai toét

sai rồi

1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)

cậu giỏi quá

đề sai rồi bạn!

Ta có :

\(1.3.5.7.....199\)

\(=\frac{1.2.3.4.5.6.7.....198.199.200}{2.4.6.....198.200}\) 

\(=\frac{\left(1.2.3.....99.100\right)\left(101.102.....200\right)}{\left(1.2.3.....99.100\right)\left(2.2.2.....2.2\right)}\)

 \(=\frac{101.102.....200}{2.2.....2}\)

\(=\frac{101}{2}.\frac{102}{2}.....\frac{200}{2}\left(đpcm\right)\)

#)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 = \(\frac{\left(1.3.5.7.....199\right).\left(2.4.6.8.....200\right)}{2.4.6.8.....200}=\frac{1.2.3.4.5.....199.200}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}=\frac{1.2.3.4.5.....199.200}{2^{100}.1.2.3.....100}=\frac{101.102.103.....200}{2^{100}}\)\(=\frac{101}{2}.\frac{102}{2}\frac{103}{2}.....\frac{200}{2}\)\( \left(ĐPCM\right)\)

1233333333333

a) dat A=1+2+2²+2³+...+2⁹⁹

2.A=2+2²+2³+2⁴+...+2¹⁰⁰

2.A-A= 2+2³+2⁴+...+2¹⁰⁰-(1+2+2²+2³+...+2⁹⁹)

A=2¹⁰⁰-1

----> 1.3.5.7...197.199<\(\frac{101.102.103....200}{2^{100}-1}\)

Dat B =1.3.5.7...197.199 

B=\(\frac{1.3.5.7....197.199...2.4.6.8....200}{2.4.6.8....200}\)

B= \(\frac{1.2.3.4.5....199.200}{2.4.6.8....200}\)

B=\(\frac{1.2.3.4.5......199.200}{2^{100}.\left(1.2.3.4...100\right)}\) ( tu 2 den 200 co 100 so hang nen duoc 2¹⁰⁰)

B =\(\frac{101.102.103....200}{2^{100}}\)

---->\(\frac{101.102.103....200}{2^{100}}

b> A= \(\frac{1.3.5.7....2499}{2.4.6.8....2500}\)  chon B=\(\frac{2.4.6.8...2500}{3.5.7.9...2501}\)

A.B = \(\frac{1.3.5.7....2499.2.4.6.8...2500}{2.4.6.8...2500.3.5.7.....2499.2501}=\frac{1}{2501}\)

Nhan xet 

\(\frac{1}{2}+\frac{1}{2}=1\)

\(\frac{2}{3}+\frac{1}{3}=1\)

vi 1/2 >1/3----> 1/2 <2/3

cm tuong tu ta se co A<B

---> A.A<A.B

---->A²<A.B

===> A² <\(\frac{1}{2501}