Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b.\left(1+\frac12\right)\left(1+\frac13\right)\left(1+\frac14\right)\ldots\left(1+\frac{1}{2023}\right)\)
\(=\frac32\cdot\frac43\cdot\frac54\cdot\ldots\cdot\frac{2024}{2023}\)
\(=\frac{3\cdot4\cdot5\cdot\ldots\cdot2024}{2\cdot3\cdot4\cdot\ldots\cdot2023}\)
\(=\frac{2024}{2}=1012\)
\(c.D=\frac{5}{6\cdot37}+\frac{1}{6\cdot43}+\frac{6}{7\cdot43}+\frac{10}{7\cdot59}\)
\(D=7\cdot\left(\frac{5}{37\cdot42}+\frac{1}{42\cdot43}+\frac{6}{43\cdot49}+\frac{10}{49\cdot59}\right)\)
\(D=7\cdot\left(\frac{1}{37}-\frac{1}{42}+\frac{1}{42}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}+\frac{1}{49}-\frac{1}{59}\right)\)
\(D=7\cdot\left(\frac{1}{37}-\frac{1}{59}\right)\)
\(D=7\cdot\frac{22}{2183}\)
\(D=\frac{154}{2183}\)
Kết quả là 2870.
Giải nhanh bằng công thức tổng bình phương:
\(1^{2} + 2^{2} + \hdots + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)
Với \(n = 20\):
\(\frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870.\)
Ta có biểu thức:
\(1\times1+2\times2+3\times3+\ldots+20\times20=1^2+2^2+3^2+\ldots+20^2\)
Đây là tổng các số chính phương từ 1 đến 20.
Áp dụng công thức tổng bình phương:
\(1^2+2^2+3^2+\ldots+n^2=\frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)
Thay \(n = 20\):
\(\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870\)
\(A=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+5^7\left(1+5+5^2\right)\)
\(=5.31+5^4.31+5^7.31=31.\left(5+5^4+5^7\right)\)chia hết cho 31
Vậy A chia 31 dư 0
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+8}\)
\(=1+\frac{1}{\left(1+2\right).3.\frac{1}{2}}+\frac{1}{\left(1+3\right).3.\frac{1}{2}}+...+\frac{1}{\left(1+8\right).8.\frac{1}{2}}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}\)
\(=1+2.\left(\frac{3-2}{2+3}+\frac{4-3}{3.4}+...+\frac{9-8}{8.9}\right)\)
\(=1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{9}\right)\)
\(=1+2.\frac{7}{18}=1+\frac{7}{9}=\frac{16}{9}\)
\(S=1+\frac{1}{\left(\frac{3.2}{2}\right)}+\frac{1}{\left(\frac{4.3}{2}\right)}+\frac{1}{\left(\frac{5.4}{2}\right)}+...+\frac{1}{\left(\frac{9.8}{2}\right)}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{9}\right)\)
\(=1+2.\frac{7}{18}\)
\(=1\frac{7}{9}\)
Chúc bn học tốt nhé!!! :)
Câu a:
M = 17/5.-31/125.1/2.10/17.-1/2^3
M = 17/17.1/[(2.5).(125.2^3)]. 31.10
M = 1.1/[10.10^3].31.10
M = 31/10^3
M = 0,031
Câu b:
N = 1/7.5/9 + 5/9.2/7 + 5/9.1/7 + 5/9.3/7
N = 5/9.(1/7 + 2/7 + 1/7+ 3/7)
N = 5/9.1
N = 5/9
\(=\frac{1}{2023}\)
\(A=\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\ldots\left(1-\frac{1}{2023}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\ldots\frac{2022}{2023}\)
\(A=\frac{1\cdot2\cdot3\cdot\ldots\cdot2022}{2\cdot3\cdot4\cdot\ldots\cdot2023}\)
\(A=\frac{1}{2023}\)
Vậy \( A=\frac{1}{2023} \)
\(A=\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\left(1-\frac16\right)\left(1-\frac17\right)\left(1-\frac18\right)\left(1-\frac19\right)\ldots\left(1-\frac{1}{2023}\right)\)
\(A=\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdot\frac56\cdot\frac67\cdot\frac78\cdot\frac89\cdot\cdot\cdot\frac{2022}{2023}\)
\(A=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot\cdot\cdot2022}{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot\cdot\cdot2023}\)
\(A=\frac{1}{2023}\)
Vậy \(A=\frac{1}{2023}\)