Cho A=1/2*3/4*5/6*..*9999/10000
Cmr A<1/100
A>1/142
Cho M=1/2*3/4*5/6*...*9999/10000 và N=2/3*4/5*6/7*...*10000/10001
a) CMR: M<N
b) CMR: M<1/100
CMR: C = 1/2 . 3/4 . 5/6 ..... 9999/10000 < 1/100
CMR
A=1/2×3/4×5/6........9999/10000<1/100
CMR C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\)
ta có :
1/2 < 2/3
2/3 <3/4
.........
9999/10000 < 10000/10001
suy ra : A2 < 1/22/33/4*****9999/1000010000/10001
suy ra : A2 < 1/10001 < 1/10000= (1/100)2
suy ra A2 < (1/100)2 . Từ đó: A < 1/100
2 là mũ 2 nha bạn
CMR :
\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}......\frac{9999}{10000}< \frac{1}{100}\)
Đặt A = (1/2)(3/4)(5/6) ... (9999/10000) (A > 0)
.Và B = (2/3)(4/5)(6/7) ... (10000/10001) (B > 0)
Ta có A.B = (1/2)(2/3)(3/4) ... (10000/10001) = 1/10001 (1)
Mặt khác :
1/2 < 2/3
3/4 < 4/5
................
................
9999/10000 < 10000/10001
Nhân tất cả vế theo vế ---> A < B ---> A² < A.B (2)
(1),(2) ---> A² < 1/10001 ---> A < căn(1/10001) < căn(1/10000) = 1/100 (đpcm)
nếu k^2=n thì ta nói căn bậc 2 của n là k(kEN)
Đặt M=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)
M<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)
M2<\(\frac{1.\left(3.5.7....9999\right)}{\left(2.4.6....10000\right)}.\frac{\left(2.4.6....10000\right)}{\left(3.5.7....9999\right).10001}\)
Bạn rút gọn đi những phần mà mình đã đóng ngoặc nha
M2<\(\frac{1}{10001}\)
M2<\(\frac{1}{10000}\)
M2<\(\left(\frac{1}{100}\right)^2\)
=> M<1/100
cho A=1/2*3/4*5/6*...*9999/10000. chứng minh rằng A <1/100
Bn tham khảo bài làm tại link này nha :
https://olm.vn/hoi-dap/detail/3377429592.html
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Câu hỏi của Thùyy Linhh - Toán lớp 6 - Học toán với OnlineMath
bài 4 : cmr :
c) C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
Ta có: \(C=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\)
\(C=\dfrac{1\cdot3\cdot5...9999}{2\cdot4\cdot6...10000}\)
Gọi \(D=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)
Mà \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};\dfrac{5}{6}< \dfrac{6}{7};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)
\(\Rightarrow\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)
\(\Rightarrow C< D\)
Ta lại có: \(C\cdot D=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\right)\)
\(\Rightarrow C\cdot D=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}...\dfrac{9999}{10000}\cdot\dfrac{10000}{10001}\)
\(\Rightarrow C\cdot D=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot6...9999\cdot10000}{2\cdot3\cdot4\cdot5\cdot6\cdot7...10000\cdot10001}\)
\(\Rightarrow C\cdot D=\dfrac{1}{10001}\)
Mà \(C< D\)
\(\Rightarrow C\cdot C< C\cdot D\)
\(\Rightarrow C\cdot C< \dfrac{1}{10001}\)
\(\Rightarrow C< \dfrac{1}{10001}\)
Mà \(\dfrac{1}{10001}< \dfrac{1}{100}\)
\(\Rightarrow C< \dfrac{1}{100}\)
Vậy \(C< \dfrac{1}{100}\)
C = \(\dfrac{1}{2}\).\(\dfrac{3}{4}\).\(\dfrac{5}{6}\)....\(\dfrac{9999}{10000}\)
C < \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}....\dfrac{10000}{10001}\)
C2 < \(\dfrac{1.\left(3.5.7...9999\right)}{\left(2.4.6...10000\right)}.\dfrac{\left(2.4.6...10000\right)}{\left(3.5.7...9999\right).10001}\)
C2 < \(\dfrac{1}{10001}\)
C2 < \(\left(\dfrac{1}{100}\right)^2\)
C < \(\dfrac{1}{100}\)
Vậy \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
Chúc bạn học tốt !
ta có : 1/2 < 2/3
2/3 <3/4
......... 9999/10000 < 10000/10001 suy ra : A2 < 1/22/33/4*****9999/1000010000/10001 suy ra : A2 < 1/10001 < 1/10000= (1/100)2 suy ra A2 < (1/100)2 . Từ đó: A < 1/100 = 0,01
cho A =1/2 . 3/4 .5/6 ....... 9999/10000
so sánh với 0,01
CMR \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}......\frac{9999}{10000}<\frac{1}{100}\)
Đặt:\(M=\frac{1}{2}\cdot\frac{3}{4}...\frac{9999}{10000}\)
\(N=\frac{2}{3}\cdot\frac{4}{5}...\frac{10000}{10001}\)
Dễ dàng nhận thấy: \(\frac{1}{2}