Trả lời:

A = 1 - 2 + 3 - 4 + 5 + ... + 53 - 54 + 55

= ( 1 + 3 + 5 + ... + 53 + 55 ) - ( 2 + 4 + ... + 54 )

\(=\left[\frac{\left(1+55\right)\cdot28}{2}\right]-\left[\frac{\left(2+54\right)\cdot27}{2}\right]\)

= 784 - 756

= 28

Ta xét trường hợp xấu nhất, bác thủy thủ Xin-bát lấy được 10 viên đỏ, 11 viên vàng.

Số kim cương ít nhất mà bác thủy thủ phải lấy để chắc chắn có đủ 3 màu là: 10 + 11 + 1 = 22 (viên)

Vậy số kim cương ít nhất mà bác thủy thủ phải lấy để chắc chắn có đủ 3 màu là 22 viên.

Ta có :

a = 2 . 3 

b =  3 . 5 . 11

=. ƯCLN( a , b ) = 3

~~Học tốt~~

Ta có:

a = 2 . 3 . 5

b = 3 . 5. 11

Ta có 2 số chung là 3 và 5

=> 3 . 5 = 15

Vậy Ư CLN (a, b) = 15

# Học tốt #

5.5.5.25 

= 5.5.5.5.5 = 5⁵

Nha bạn !!!

Viết gọn các tích sau bằng cách dùng lũy thừa C,5.5.5.25  là 5\(^5\)

chúc bn hok tốt

\(\frac{-5}{18}\times\frac{7}{11}-\frac{7}{18}\times\frac{13}{11}+1\frac{7}{11}.\)

= \(\frac{-5}{18}\times\frac{7}{11}-\frac{7}{18}\times\frac{13}{11}+\frac{18}{11}\)

= \(-\frac{35}{198}-\frac{91}{198}+\frac{324}{198}\)

= 1

Mong mọi người trả lời nhanh nhé, khẩn cấp!!!

Ta có \(\frac{2021}{1}+\frac{2020}{2}+\frac{2019}{3}+...+\frac{2}{2020}+\frac{1}{2021}\)

\(=1+\left(\frac{2020}{2}+1\right)+\left(\frac{2019}{3}+1\right)+...+\left(\frac{2}{2020}+1\right)+\left(\frac{1}{2021}+1\right)\)

\(=\frac{2022}{2022}+\frac{2022}{2}+\frac{2022}{3}+...+\frac{2022}{2020}+\frac{2022}{2021}\)

\(=\frac{2022}{2}+\frac{2022}{3}+...+\frac{2022}{2020}+\frac{2022}{2021}+\frac{2022}{2022}\)

\(=2022\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}\right)\)

Khi đó M = \(\frac{2022\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2021}+\frac{1}{2022}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2021}+\frac{1}{2022}}=2022\)

Trả lời:

Ta có: \(\frac{2021}{1}+\frac{2020}{2}+\frac{2019}{3}+...+\frac{2}{2020}+\frac{1}{2021}\)

\(=\left(1+1+...+1+1+1\right)+\frac{2020}{2}+\frac{2019}{3}+...+\frac{2}{2020}+\frac{1}{2021}\)

\(=\left(\frac{2020}{2}+1\right)+\left(\frac{2019}{3}+1\right)+...+\left(\frac{2}{2020}+1\right)+\left(\frac{1}{2021}+1\right)+1\)

\(=\frac{2022}{2}+\frac{2022}{3}+...+\frac{2022}{2020}+\frac{2022}{2021}+\frac{2022}{2022}\)

\(=2022\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}\right)\)

Khi đó: \(M=\frac{2022\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}}=2022\)

Ta có \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2014.2015}\)

\(\)\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2015}=\frac{1}{5}-\frac{1}{2015}=\frac{402}{2015}\)

=> \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{402}{2015}\)(1)

Lại có \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2013.2014}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2014}=\frac{1}{4}-\frac{1}{2014}< \frac{1}{4}\)

=> \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}\left(2\right)\) 

Từ (1) và (2) => \(\frac{402}{2015}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}\)

Trả lời:

Ta có: \(\frac{1}{5^2}>\frac{1}{5\cdot6};\frac{1}{6^2}>\frac{1}{6\cdot7};\frac{1}{7^2}>\frac{1}{7\cdot8};...;\frac{1}{2014^2}>\frac{1}{2014\cdot2015}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2014\cdot2015}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{1}{5}-\frac{1}{2015}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}>\frac{402}{2015}\)(1)

Ta có: \(\frac{1}{5^2}< \frac{1}{4\cdot5};\frac{1}{6^2}< \frac{1}{5\cdot6};\frac{1}{7^2}< \frac{1}{6\cdot7};...;\frac{1}{2014^2}< \frac{1}{2013\cdot2014}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2013\cdot2014}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}-\frac{1}{2014}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{402}{2015}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2014^2}< \frac{1}{4}\)

Ta có : \(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)

<=> \(\frac{xy-4}{4y}=\frac{1}{2}\)

<=> 2(xy - 4) = 4y

<=> xy - 4 = 2y

<=> xy - 2y = 4

<=> y(x - 2) = 4

Lập bảng xét các trường hợp 

Vậy các cặp (x;y) thỏa mãn (3;4) ; (6;1) ; (1;-4) ; (-2 ; -1) ; (4;2) ; (0;-2)