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\(A=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+5^7\left(1+5+5^2\right)\)
\(=5.31+5^4.31+5^7.31=31.\left(5+5^4+5^7\right)\)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!!! :)
A= 7/8:(4/18-1/18)+7/8:(1/36-15/36)
=7/8:1/6+7/8:(-7/18)
=7/8:(1/6+-7/18)=7/8:(3/18+-7/18)=7/8:(-2/9)=-63/18=-7/2
\(=\frac{1}{2023}\)
\(A=\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\ldots\left(1-\frac{1}{2023}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\ldots\frac{2022}{2023}\)
\(A=\frac{1\cdot2\cdot3\cdot\ldots\cdot2022}{2\cdot3\cdot4\cdot\ldots\cdot2023}\)
\(A=\frac{1}{2023}\)
Vậy \( A=\frac{1}{2023} \)