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\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{624}{625}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{24.26}{25.25}\)
\(=\frac{1.2.3....24}{2.3.4....25}\cdot\frac{3.4.5....26}{2.3.4....25}\)
\(=\frac{1}{25}\cdot\frac{26}{2}=\frac{26}{50}=\frac{13}{25}\)
\(\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{8}\right)\cdot\left(1+\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot\cdot\cdot\cdot\frac{10000}{9999}\)
\(=\frac{2.2}{1.3}\cdot\frac{3.3}{2.4}\cdot\frac{4.4}{3.5}\cdot\cdot\cdot\cdot\frac{100.100}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}\cdot\frac{2.3.4...100}{3.4.5...101}\)
\(=\frac{100}{1}\cdot\frac{2}{101}=\frac{200}{101}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4...100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)
A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )
3/4.8/9.15/16.....624/625
=(1.3)/(2.2).(2.4)/(3.3).(3.5)/(4.4)...(24.26)/(25.25)
=(1.2.3....24).(3.4.5....26)/(2.3.4...25).(2.3.4...25)
=26/25.2
=26/50
=13/25
Đặt\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)
Vì\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)\(>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(mỗi cái trong ngoặc là một trăm phân số)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\left(\frac{1}{200}\right).100+\left(\frac{1}{300}\right).100\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A>\frac{5}{6}\)
Mà 5/6>2/3=>A>2/3
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Vì \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{103}+.....\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}\right)\)
Tự làm tiếp nhé !!!
gọi số phải tìm là A A
=(1.3).(2.4).(3.5)...(99.101)/
(2².3².4²...100²)
=(1.2.3...99).(3.4.5...101)/
[(1.2.3.4...100)(2.3.4...100)]
=101/(100.2)=101/200
Mày hay nhờ mai tao méc thầy
tự làm đi