E = 2/5.10 + 2/10.15 + ... + 2/35.40

E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)

E = 2/5.(1/5 - 1/40)

E = 2/5.7/40

E = 7/100

E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

   = \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\frac{7}{40}\)

   = \(\frac{7}{100}\)

4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2008.2010

= 2.(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2008 - 1/2010)

= 2.(1/2 - 1/2010)

= 2.502/1005

= 1004/1005

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2.\frac{502}{1005}\)

\(=\frac{1004}{1005}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{2}{99}-\frac{2}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{303}{303}-\frac{3}{303}=\frac{300}{303}\)

\(=\frac{100}{101}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}\)

\(=\frac{98}{303}\)

a) Ta có : aaa = a . 111 = a . 37 . 3 \(⋮\)37

=> aaa \(⋮\)37 (đpcm)

b) Ta có: aaaaaa = a . 111111 = a . 37 . 3003 \(⋮\)37

=> aaaaaa \(⋮\)37 (đpcm)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

1/1*2+1/2*3+1/3*4+......+1/99*100+1/100*101            Bài tương tự

= 1-1/2 + 1/2-1/3+...+1/99-100+1/00-1/101

= 1-1/101

=100/101

tk nha

So sánh : 

a ) 31^11 và 17^14

31^11 < 32^11= (2⁵)¹¹ = 2^55

=> 31^11 < 2^55

17^14>16^14=(2⁴)¹⁴ = 2^56

=>17^14>2^56

=>31^11 < 2^55 < 2^56 < 17^14

=>31^11 < 17^14

b ) 3^500 và 7^300

3^500 = ( 3⁵)¹⁰⁰ = 243¹⁰⁰

7^300 = ( 7³)¹⁰⁰ = 343¹⁰⁰

=> 243¹⁰⁰ < 343¹⁰⁰

=> 3^500 < 7^300

Tìm x : 

a ) 2^x . 4 = 128

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

b ) 2^x . 2² = ( 2³)² = 64

=> 2^x = 64 : 2² = 16

=> 2^x = 2⁴

=> x = 4

Bài cuối bạn tham khảo tại : Câu hỏi của Linh Phan - Toán lớp 6 - Học toán với OnlineMath

Link : https://olm.vn/hoi-dap/detail/198524999512.html

MONG NM TRẢ LỜI GIÚP CHO!!!