\(M=1+2.\left(\dfrac{3}{2.5}+\dfrac{5}{5.10}+\dfrac{7}{10.17}+\dfrac{9}{17.26}\right)\)

\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{26}\right)\)

\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{26}\right)\)

\(=1+1-\dfrac{1}{13}=\dfrac{25}{13}\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)

Đặt \(B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

=>\(B=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)

=>\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{50}\right)=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2\)

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

\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]\)

Ta có:

\(\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2.2}< \dfrac{1}{2.1}=\dfrac{2-1}{2.1}=\dfrac{2}{2.1}-\dfrac{1}{2.1}=1-\dfrac{1}{2}\)

\(\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3.3}< \dfrac{1}{3.2}=\dfrac{3-2}{3.2}=\dfrac{3}{3.2}-\dfrac{2}{3.2}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\left(\dfrac{1}{50}\right)^2=\dfrac{1}{50.50}< \dfrac{1}{50.49}=\dfrac{50-49}{50.49}=\dfrac{50}{50.49}-\dfrac{49}{50.49}=\dfrac{1}{49}-\dfrac{1}{50}\)

Khi đó

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

\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]< \dfrac{1}{4}.2=\dfrac{1}{2}\)

Vậy \(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2< \dfrac{1}{2}\left(đpcm\right)\)

Tick cho mk nha :>>

1 C

2 B

3 A

4 D

5 B

6 B

7 D

8 C

9 D

10 D

11 C

12 A

13 B

14 D

15 B

đâu buôi

@ Bùi Thành An 

không nói linh tinh và lăng mạ nhé

4106 có các ước là 1, 2, 2053, 4106

4106 có các ước là 1, 2, 2053, 4106

When will the students go camping?

A = \(\dfrac{n+4}{3n+5}\) (n \(\ne\) - \(\dfrac{5}{3}\))

A \(\in\) Z ⇔ n + 4 \(⋮\) 3n + 5

          3(n + 4) ⋮ 3n + 5

          3n + 12 ⋮ 3n + 5

      3n + 5 + 7 ⋮ 3n + 5

                    7 ⋮ 3n + 5

            3n + 5 \(\in\) Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có: 

Theo bảng trên ta có n \(\in\) {-4; - 2}

Kết luận A = \(\dfrac{n+4}{3n+5}\) có giá trị nguyên khi và chi khi n \(\in\) {- 4; - 2}

Ta có:

Để \(\dfrac{n+4}{3n+5}\) đạt giá trị nguyên thì \(\left(n+4\right)⋮\left(3n+5\right)\)

\(\Rightarrow3\left(n+4\right)⋮3n+5\)

\(\Rightarrow\left(3n+5+7\right)⋮\left(3n+5\right)\)

\(\Rightarrow7⋮\left(3n+5\right)\)

\(\Rightarrow3n+5\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow3n\in\left\{-12;-6;-4;2\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-\dfrac{4}{3};\dfrac{2}{3}\right\}\)

\(-\dfrac{3}{7}-\dfrac{1}{4}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{5}{4}\\ =\dfrac{3}{7}.\left(-1-\dfrac{1}{4}+\dfrac{5}{4}\right)\\ =\dfrac{3}{7}.0\\ =0\)

\(\dfrac{-3}{7}-\dfrac{1}{4}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{5}{4}\)

=\(\dfrac{-3}{7}-\dfrac{3}{28}+\dfrac{15}{28}\)

=\(\dfrac{-15}{28}+\dfrac{15}{28}\)

=\(0\)

\(#LilyVo\)