Ta có:

65 × 111 - 13 × 15 × 37

= 5 × 13 × 3 × 37 - 13 × 3 × 5 × 37

= 0

Vì 0 nhân với bất kì số nào cũng = 0 nên biểu thức trên = 0

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(65.111-13.15.37\right)\)

\(\left(1+2+3+...100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(13.5.111-13.15.37\right)\)

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(13.15.37-13.15.37\right)\)

\(=0\)

c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))

= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)

= (17 - 6)  - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))

= 11  - \(\dfrac{15}{17}\)+ 0

=    \(\dfrac{172}{17}\)

b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)

= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)

= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))

= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))

= 250 +  \(\dfrac{193}{140}\)

= 250\(\dfrac{193}{140}\)

6.B=1.3.6+3.5.6+5.7.6+...+95.97.6+97.99.6

6.B=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+95.97.(99-93)+97.99(101-95)

6.B=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+95.97.99-93.95.97+97.99.101-95.97.99=1.3+97.99.101

B=(3+97.99.101)/6

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

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

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)

\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)

\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)

Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
 

Bạn tham khảo nhé!

Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101

A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)

A = (1² + 3² + 5² + … + 97² + 99²) + 2.(1 + 3 + 5 + … + 97 + 99).

Đặt B = 1² + 3² + 5² + … + 99²

=> B = (1² + 2² + 3² + 4² + … + 100²) – 2².(1² + 2² + 3² + 4² + … + 50²)

Tính dãy tổng quát C = 1² + 2² + 3² + … + n²

C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]

C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)

C =  = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6

Áp dụng vào B ta được:

B = 100.101.201 : 6 – 4.50.51.101 : 6  = 166650

=> A = 166650 + 2.(1 + 99).50 : 2

=> A = 166650 + 5000 = 172650.

Đ/s: A = 172650.

  A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)

2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)

2A= 1-\(\dfrac{1}{99}\)

2A= \(\dfrac{98}{99}\)

  A= \(\dfrac{98}{99}\) : 2

A=\(\dfrac{49}{99}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)