\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)

Đặt \(A=2.4+4.6+...+50.52\)

\(\Rightarrow6A=2.4.6+4.6.6+...+50.52.6\)

\(=2.4.6+4.6.\left(8-2\right)+....+50.52.\left(54-48\right)\)

\(=2.4.5+4.6.8-2.4.6+...+50.52.54-48.50.52\)

\(=50.52.54\)

\(\Rightarrow A=\frac{50.52.54}{6}=23400\)

\(\text{Đặt A = 2.4 + 4.6 + ... + 50.52}\)

\(\text{⇒6A = 2.4.6 + 4.6.6 + ... + 50.52.6}\)

\(\text{6A= 2.4.6 + 4.6. 8 − 2 + .... + 50.52. 54 − 48 }\)

\(\text{6A= 2.4.5 + 4.6.8 − 2.4.6 + ... + 50.52.54 − 48.50.52}\)

\(\text{6A=50.52.54 }\)

\(\Rightarrow A=\)\(\frac{\text{50.52.54 }}{6}\)\(=23400\)

6A=2.4.6+4.6.(8-2)+.........+50.52.(54-48)

6A=2.4.6+4.6.8-2.4.6+............+50.52.54-48.50.52

6A=50.52.54

A=50.52.54:6

A=23400

Bài 1:

Kiểm tra lại đề đi cậu

Bài 2:

5^x + 5^x+1 = 750  

5^x + 5^x.5=750 

5^x . ( 1 + 5 ) = 750

5^x . 6           = 750

5^x                    = 750 : 6  

5^x               = 125

5^x               = 5³

=> x = 3

Đổi dấu bài 1 thành như này nha mấy bạn :

 trừ > cộng > trừ > cộng > ... > cộng > trừ

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)

                                                   \(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

Đặt A = 2/2.4+2/4.6+....+2/2010.2012

         = 1/2 - 1/4 + 1/4 - 1/6 + ...... + 1/2010 - 1/2012

        = 1/2 - 1/2012

        = 1005/2012

2/2.4+2/4.6+...+2/2010.2012

= 1/2-1/4+1/4-1/6+...+1/2010-1/2012

= 1/2-1/2012

= 1006/2012-1/2012

= 1005/2012

1/2-1/4+1/4-1/6+....+1/2010-1/2012

2

1/2-1/2012

2

1005/2024

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2004.2006}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2004}-\frac{1}{2006}\)

\(=\frac{1}{2}-\frac{1}{2006}\)

\(=\frac{1003}{2006}-\frac{1}{2006}\)

\(=\frac{1002}{2006}\)

\(=\frac{501}{1003}\)