Ta có \(k^2>k^2-1=\left(k+1\right)\left(k-1\right)\)
Áp dung vào bài toán ta được
\(A=\frac{1}{2}.\frac{3}{4}...\frac{199}{200}=\frac{1.3...199}{2.4...200}\)
\(\Rightarrow A^2=\frac{1^2.3^2...199^2}{2^2.4^2...200^2}< \frac{1^2.3^2...199^2}{1.3.3.5...199.201}=\frac{1^2.3^2...199^2}{1.3^2.5^2...199^2.201}=\frac{1}{201}\)
Vậy \(A^2< \frac{1}{201}\)
Cho \(A=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{199}{200}\)và chứng minh \(A^2< \frac{1}{201}\)
Rút gọn:
\(A=\frac{200+\frac{199}{2}+\frac{198}{3}+...+\frac{2}{199}+\frac{1}{200}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{200}+\frac{100}{201}}\)
\(a.\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Cho C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\) Cm C2 < \(\frac{1}{201}\)
( 2 cách nha )
Cho \(S=\frac{1}{2}+\frac{3}{4}+\frac{5}{6}+...+\frac{199}{200}\)Chứng minh rằng : \(S^2<\frac{1}{201}\)
1. thực hiện phép tính
a, 36+3.(4-12) b. \(2-\frac{3}{4}\)
c.\(\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\) d.\((\frac{12}{199}-\frac{23}{200}+\frac{34}{201})-(\frac{1}{2}-\frac{1}{3}-\frac{1}{6})\)
cho A =\(\frac{1.3.5......199}{2.4.6.......200}\)CMR :A2<\(\frac{1}{201}\)
Chứng minh rằng
a)\(\frac{5}{8}<\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}<\frac{3}{4}\)
b)\(\frac{1}{4}+\frac{1}{6}+\frac{1}{16}+...+\frac{1}{10000}<\frac{3}{4}\)
c)(\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\))2 <\(\frac{1}{201}\)
d)\(50<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{2^{100}-1}<100\)
1. Tính nhanh
a, \(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\)
b, \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c,\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
