<1/25+1/25+...+1/25+1/40+1/40+1/40+...+1/40

=1/25.15+1/40.15=15/25+15/40=3/5+3/8=39/40

đặt vế trái là A ta có A<1/25.16+1/45.14=214/225=1-11/225 39/40=1-1/40 ta có 1.225<11.40=>1/40<11/225=>1-1/40>1-11/225=>214/225<39/40 mà A<214/225=>A<39/40(đpcm nhớ k cho mk nha

=1/25+1/54

=79/1350

vi 79/1350<39/40

nen 1/25+1/26+1/27+...1/54<39/40

Ta có : 1/25 + 1/26 + 1/27 +.....+1/39 < 1/25 + 1/25 + .....+1/25 = 15/25 = 3/5

1/40+1/41 +.....+1/54 < 1/40 + 1/40 +....+1/40 = 15/40 = 3/8

=> A = 1/25 + 1/26 + 1/27 +.......+1/54 < 3/5 + 3/8 = 39/40

=> A < 39/40   (đpcm)

Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

Khi đó : \(\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

\(=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)=1\) (đpcm)

Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

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

Khi đó \(\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}}=\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}=1\left(\text{đpcm}\right)\)

1. Bài toán 46

2. Chưa tìm ra

1: vì 9>5 nên làm tròn thành 1

2: chịu

Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)(đpcm)

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)   (đpcm)

Giải:

\(\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\) 

Ta có:

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\) 

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\) 

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)\) 

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\left(đpcm\right)\) 

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

=1/1-1/2+1/3-1/4+...+1/49-1/50

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

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

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

Tự hỏi tự trả lời

b1/A=25/1.6+25/6.11+25/11.16+....+25/41.46

=5.(5/1.6+5/6.11+5/11.16+...+5/41.46)

=5.(1/1-1/6+1/6-1/11+1/11-1/16+....+1/41-1/46)

=5.(1/1-1/46)

=5.45/46

=225/46