Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)
Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)

                =  (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <  1 - 1 / 2 + 1 / 3 = 5 / 6 (2)
Từ (1) và (2) => 7 / 12 < A < 5 / 6

@Lê Nguyên Hạo

nhận dc thông báo ko

a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦ 
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) = 
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 < 
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥ 
♦, ♥ => 7 / 12 < A < 5 / 6

b)ta có:

1/1.2+1/3.4+1/5.6+...+1/49.50

=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)

=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2

=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)

=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50

hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75},\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{100}-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)

\(\Rightarrow A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(1\right)\)

Lại có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{25}{50}=\frac{1}{2}\)

\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

\(\Rightarrow A< \frac{1}{2}+\frac{1}{3}=\frac{5}{6}\left(2\right)\)

Từ (1) và (2) => \(\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right)\)

\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)

\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)

Bài giải

a)

Xét phân số 1/[n(n+1)] với n lẻ.

Ta có:
1/[n(n+1)] = 1/n - 1/(n+1)

Vậy:
1/(1.2) = 1/1 - 1/2
1/(3.4) = 1/3 - 1/4
1/(5.6) = 1/5 - 1/6
...
1/(49.50) = 1/49 - 1/50

Suy ra:

1/(1.2) + 1/(3.4) + 1/(5.6) + ... + 1/(49.50)
= (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + ... + (1/49 - 1/50)

= (1 + 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + 1/6 + ... + 1/50)

Mà:
(1 + 1/2 + 1/3 + ... + 1/50) - (1/2 + 1/4 + 1/6 + ... + 1/50)
= 1 + 1/3 + 1/5 + ... + 1/49

Nên:
1/(1.2) + 1/(3.4) + ... + 1/(49.50)
= (1 + 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + ... + 1/50)

= (1 + 1/2 + 1/3 + ... + 1/50) - 2(1/2 + 1/4 + ... + 1/50)

= (1 + 1/2 + 1/3 + ... + 1/50) - (1 + 1/2 + ... + 1/25)/?

Sau khi sắp xếp lại, ta thu được:

= 1/26 + 1/27 + ... + 1/50

Đpcm.

b)

Gọi A = 1/(1.2) + 1/(3.4) + 1/(5.6) + ... + 1/(99.100)

Theo kết quả a), ta có:

A = 1/51 + 1/52 + 1/53 + ... + 1/100

Có tất cả 50 số hạng.

Mà:
7/12 = 0,5833...
5/6 = 0,8333...
1/2 = 0,5
50/51 ≈ 0,9803

Vậy ta có:
7/12 < A < 5/6

Đpcm.

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

Vì \(\frac{7}{12}< \frac{99}{100}< \frac{5}{6}\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)               ĐPCM

( Bài này ko ai lm thì t lm cho )