CHO A =1/2.3/4.5/6<9999/10000

mình sorry, mih nhầm

đáp án phải là .................10 123 499

Bây gio thì chac chan dung

Nhớ bấm ***k cho hinh nha

giup minh nha

132442+224*22622-9999+222*22224=10123499

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

B< 1+(1/1.2+1/2.3+...+1/62.63)

B<1+(1-1/2+1/2-1/3+...+1/62-1/63)

B<1+1-1/63

B<2-1/63

B<6-3/189

mà 6-3/189<6 

Vậy B<6

b, gọi D=2/3.4/5....10000/10001

Ta có: 1/2<2/3     3/4<4/5      .. .....      9999/10000<10000/10001

=> C<D                 1

C.D=1/2.3.4.....9999/10000.2/3.4/5...10000/10001

C.D=1/10001       2

Từ 1 : C<D => C.C<C.D<1/10001

                   =>C^2<1/10001<1/10000

                   =>C^2<(1/100)^2

Vậy C<1/100 (đpcm)

Ta cóC= \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}......\dfrac{9999}{10000}\)

Đặt A = \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{10000}{10001}\)

Khi đó AC = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}\)= \(\dfrac{1}{10001}\)

Do \(\dfrac{1}{2}< \dfrac{2}{3}\)

\(\dfrac{3}{4}< \dfrac{4}{5}\)

.............

\(\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

=> C<A=>C²<CA hay C²< \(\dfrac{1}{10001}\) , mà \(\dfrac{1}{10001}\)<\(\dfrac{1}{10000}\)=> C²< \(\dfrac{1}{10000}\)

^{Khi đó C < \(\sqrt{\dfrac{1}{10000}}\)}hay C< \(\dfrac{1}{100}\)( đpcm )

a, Ta có : \(\dfrac{1}{2^2}=\dfrac{1}{4};\dfrac{1}{3^2}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{4^2}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(...\dfrac{1}{100^2}< \dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}< 2\)

@Nguyễn Khanh

b, 1 = 1
1/2 + 1/3 = 1/(1 + 1) + 1/(1 + 2) < 2/(1 + 1) = 2/2 = 1
1/4 + 1/5 + 1/6 + 1/7 = 1/(3 + 1) + 1/(3 + 2) + 1/(3 + 3) + 1/(3 + 4) < 4/(3 + 1) = 4/4 = 1
1/8 + 1/9 + ... + 1/15 = 1/(7 + 1) + 1/(7 + 2) + ... + 1/(7 + 8) < 8/(7 + 1) = 8/8 = 1
1/16 + 1/17 + ... + 1/31 = 1/(15 + 1) + 1/(15 + 2) + ... + 1/(15 + 16) < 16/(15 + 1) = 16/16 = 1
1/32 + 1/33 + ... + 1/63 = 1/(31 + 1) + 1/(31 + 2) + ... + 1/(31 + 32) < 32/(31 + 1) = 32/32 = 1
=> 1 + 1/2 + 1/3 + 1/4 + ... + 1/64 < 1 + 1 + 1 + 1 + 1 + 1
=> 1 + 1/2 + 1/3 + 1/4 + ... + 1/64 < 6 (đpcm)
@Nguyễn Khanh

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 theo vế ---> A < B ---> A² < A.B (2)
Từ (1),(2) => A² < 1/10001 => A < \(\sqrt{\dfrac{1}{10001}}\) < \(\sqrt{\dfrac{1}{10000}}\) = 1/100 (đpcm)

@Nguyễn Khanh

 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)

Ta có C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)

Gọi D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\)

Mà \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{9999}{10000}< \frac{10000}{10001}\)

=> C<D

Lại có C.D = \(\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right)\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)

C.D = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{9999}{10000}.\frac{10000}{10001}\)

C.D = \(\frac{1}{10001}\)

Vì C<D

=> C.C < C.D

hay  C.C <\(\frac{1}{10001}\)

=> C < \(\frac{1}{10001}< \frac{1}{100}\)(đpcm)

1/2.3/4.5/6....9999/10000

=1/2