P=1/2.3/4.5/6.....399/400.Chung to rang P nho hon1/20
cho biểu thức :
p=1/2.3/4.5/6....399/400
chứng minh:p<1/20
Đặt \(Q=\frac{2}{3}.\frac{4}{6}.\frac{6}{7}....\frac{400}{401}\)
Áp dụng tính chất \(\frac{a}{b}< \frac{a+m}{b+m}\left(a,b,m\inℕ^∗\right)\)ta có :
\(\frac{1}{2}< \frac{1+1}{2+1}=\frac{2}{3}\)
\(\frac{2}{3}< \frac{2+1}{3+1}=\frac{3}{4}\)
...
\(\frac{399}{400}< \frac{399+1}{400+1}=\frac{400}{401}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{399}{400}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{400}{401}\)
Hay \(P< Q\)
\(\Rightarrow P^2< P.Q\)
\(P^2< \frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{399}{400}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{400}{401}\)
\(P^2< \frac{1.2.3.4.....400}{2.3.4.5.....401}\)
\(P^2< \frac{1}{401}< \frac{1}{400}< \left(\frac{1}{20}\right)^2\)
Vì \(P\)và \(\frac{1}{2}\)có cùng dấu
\(\Rightarrow P< \frac{1}{2}\)
Hk tốt
p=1/2.3/4.5/6......399/400
=>p<1/2.2/4.4/6....398/400
p<1.2.4.....398/2.4.6....400
rut gon dc p<1/400<1/20
vay p < 1/20
Cho:
P=1/2.3/4.5/6.....399/400.Chứng tỏ P<1/20
Mình đố các bạn thử nha
Đặt Q =\(\frac{2}{3}\) . \(\frac{4}{5}\) . \(\frac{6}{7}\) . \(\frac{8}{9}\) ......\(\frac{400}{401}\)
Mà P = \(\frac{1}{2}\) . \(\frac{3}{4}\) . \(\frac{5}{6}\) . \(\frac{7}{8}\) .......\(\frac{399}{400}\)
➜ P < Q
Ta có : P . Q = 1/2.2/3.3/4.4/5.......399/400.400/401
=\(\frac{1.2.3.....399.400}{2.3.4.....400.401}\)
= \(\frac{1}{401}\) < 1/400 ( \(\frac{1}{20}\) )
Mà P2 < P.Q < ( 1 /20 )2
⇔ P < \(\frac{1}{20}\) ( đpcm )
Cho:
P=1/2.3/4.5/6.....399/400.Chứng tỏ P<1/20
Mình đố các bạn thử nha
\(P< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{400}{401}\)
\(P^2< \frac{1.2.3...400}{2.3.4...401}=\frac{1}{401}< \frac{1}{400}\)
\(\Rightarrow P< \frac{1}{20}\)
Chung to rang: 1/201+1/202+...+1/399+1/400>1/2
Các phân số 1/201; 1/202;....;1/399 đều lớn hơn 1/400 nên 1/201+1/202+...+1/399+1/400>1/400 . 200 = 1/2
chung to rang: 1/201+1/202+...+1/399+1/400 > 1/2
\(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{400}>\frac{1}{400}.200=\frac{200}{400}=\frac{1}{2}\)
=> điều phải chứng minh.
Đinh Tuấn Việt thì lúc nào cũng giỏi rồi
cho A=1/2.3/4.5/6.7/8.....199/200
chung to rang A2<1/201
Cho M = 1/2.3/4.5/6................99/100
N = 2/3.4/5.6/7...............100/101
a,Chung minh rang M<N
b,Tinh M.N
c,Chung minh rang M<1/10
minh can cach lam
ai nhanh minh tich cho va ket ban nua
chung to rang tong sau nho hon 2 A=6/17+9/16+11/34
cho a =1/2.3/4.5/6.....99/100 chung minh rang1/15<a<1/10
cho a =1/2.3/4.5/6.....99/100.Chứng minh rằng:1/15<a<1/10.
ta co a < 2/3.4/5.....100/101
nhan hai ve cho a ta co
a^2 <2/3.4/5...100/101.1/2.3/4.5/6...99/100
a^2<1/101 <1/100
a< can 1/100 a <1/10.
Cm tương tự ta dc a>1/15.
Bn cx có thể kham khảo bài làm khác là:https://diendan.hocmai.vn/threads/toan-6-cmr-a-1-10-va-a-1-15.223994/
vì: Ta có a:1/2=3/4.5/6.7/8...99/100
=> a<3/4.5/6..99/100