Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

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

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

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

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

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

con khỉ tao đéo b

= 1 -1/2 + 1/2 - 1/3 + 1/3 -...-1/60

= 1-1/60=59/60

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

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

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

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.

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

                                                                     \(=1-\frac{1}{6}\)

                                                                     \(=\frac{5}{6}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1-1/6

=5/6

1/1 - 1/2+1/2 -1/3+1/3 - 1/4+1/4 - 1/5+1/5 - 1/6

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

=1 - 1/6

=6/6 - 1/6

=5/6

mk bít lm cách lớp 5, vừa học

Cần ko bn

Đặt Q = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{997.998}+\frac{1}{999.1000}\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{997.999}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{997}-\frac{1}{999}\)

\(2A=1-\frac{1}{999}\)

\(2A=\frac{998}{999}\)

\(\Leftrightarrow A=\frac{499}{999}\)

Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{998.1000}\)

\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{998}-\frac{1}{1000}\)

\(2B=\frac{1}{2}-\frac{1}{1000}\)

\(B=\frac{499}{1000}\)

Vậy Q = A + B = \(\frac{499}{999}+\frac{499}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{999.1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}=\frac{999}{1000}\)

Đặt 

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{999.1000}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(\Leftrightarrow A=1-\frac{1}{1000}\)

\(\Leftrightarrow A=\frac{999}{1000}\)

Đặt:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

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

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

bạn viết sai đề rồi

Đặt biểu thức là A.

A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

A=\(\frac{1}{1}-\frac{1}{1000}\)

A=\(\frac{999}{1000}\)

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

\(A=1-\frac{1}{6}=\frac{5}{6}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)

ui cí này e chưa học

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}\)

\(=\frac{5}{6}\)