Số tròn triệu đó có dạng: \(\overline{a000000}\left(0\le a\le9\right)\)

Khi chia cho 10000

\(\overline{a000000}:10000=\overline{a00}\)

\(\rightarrow\)Chữ số hàng chục và chữ số hàng đơn vị đều là 0

Vậy cả hai bạn Đào và Mận đều sai.

Nữa chu vi hình chữ nhật :

`120:2=60(cm)`

Hiệu chiều dài và chiều rộng :

`4+4=8(cm)`

Chiều dài hình chữ nhật:

`(60+8)/2=34(cm)`

Chiều rộng hình chữ nhật :

`34 - 8 = 26(cm)`

Đ/s...

j

\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}+\dfrac{1}{168}+\dfrac{1}{224}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}+\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{16}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{16}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{7}{16}=\dfrac{7}{32}\)

kết bn ko HIỀN THƯƠNG

Số lớn :

`(2015 + 199)/2 = 1107`

Số bé :

`2015 - 1107 = 908`

Vậy .....

H

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{50}{51}=\dfrac{1}{51}=\dfrac{3}{153}< \dfrac{3}{150}< \dfrac{13}{150}\)

=> A<B

\(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+...+\dfrac{1}{21\cdot23}+\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{21\cdot23}+\dfrac{2}{23\cdot25}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{21}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{25}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{25}=\dfrac{2}{25}\)

9009 : 33

\(\dfrac{3}{2}\times5+\dfrac{4}{5}\times9+\dfrac{5}{9}\times14\)

\(=\dfrac{3\times5}{2}+\dfrac{4\times9}{5}+\dfrac{5\times14}{9}\)

\(=\dfrac{15}{2}+\dfrac{36}{5}+\dfrac{70}{9}=\dfrac{2023}{90}\)

\(\dfrac{24}{7}\) giờ

\(=\dfrac{21}{7}\) giờ \(+\dfrac{3}{7}\) giờ

\(=3\) giờ \(\dfrac{180}{7}\) phút

Vậy \(\dfrac{24}{7}\) giờ \(=3\) giờ \(\dfrac{180}{7}\) phút.

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{50}{51}=\dfrac{1}{51}=\dfrac{3}{153}< \dfrac{3}{150}< \dfrac{13}{150}\)

\(\Rightarrow A< B\)

Ta có 

\(A=\dfrac{1}{1\cdot2}\cdot\dfrac{4}{2\cdot3}\cdot\dfrac{9}{3\cdot4}\cdot\dfrac{16}{4\cdot5}..\cdot\dfrac{2500}{50\cdot51}\)

\(A=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}+...+\dfrac{50^2}{50\cdot51}\)

\(A=\dfrac{1^2\cdot2^2\cdot3^2\cdot...\cdot50^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot..\cdot50\cdot51}\)

\(A=\dfrac{1^2\cdot2^2\cdot3^2\cdot...\cdot50^2}{1\cdot2^2\cdot3^2\cdot4^2\cdot...\cdot50^2\cdot51}\)\(=\dfrac{1}{51}\)

Có \(A=\dfrac{1}{51}=\dfrac{13}{663}< B=\dfrac{13}{150}\)