Cho \(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}.\frac{8}{7}.......\frac{200}{199}\)
CMR: 201<S2<400
Giải đúng và chi tiết mk **** cho
Cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\).CMR \(A^2\)<\(\frac{1}{201}\)
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 \(A^2< \frac{1}{201}\)
CMR
\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)
Có C^2 < 1/201
C = 1/200
=> C^2 = 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
K nhé!
Ta rút gọn C = 1/200
=> C^2 = 1/400
Mà 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
Ai k mk mk k lại !!
Ta rút gọn C = 1/200
=> C^2 = 1/400
Mà 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
Cho \(S=\frac{1}{2}+\frac{3}{4}+\frac{5}{6}+...+\frac{199}{200}\)Chứng minh rằng : \(S^2<\frac{1}{201}\)
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,\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
Tính nhanh:
a,\(\frac{7}{13}\cdot\frac{7}{15}-\frac{5}{12}\cdot\frac{21}{39}+\frac{49}{91}\cdot\frac{8}{15}\)
b,\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Cho \(A=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{199}{200}\)và chứng minh \(A^2< \frac{1}{201}\)
ta có 1/2<2/3 ; 3/4<4/5;5/6<6/7;...;199/200<200/201
suy ra A^2=1/2^2*3/4^2*5/6^2*...*199/200^2<1/2*2/3*3/4*4/5*5/6*6/7*...*199/200/200/201
suy ra A^2<1/201(đpcm)
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow A.A< A.\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\)(làm phần trc như Sakuraba Laura nhá)
Cho C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\) Cm C2 < \(\frac{1}{201}\)
( 2 cách nha )
cho \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\)
chứng minh :\(C^2
Cho A=\(\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}CMR:14< A< 20\)