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\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
Ta có :
\(A=1+2+2^2+2^3+...+2^{2009}+2^{2010}\)
\(=1+\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=1+7+2^4\left(2+2^2+2^3\right)+...+2^{2008}\left(2+2^2+2^3\right)\)
\(=1+7+2^4.7+2^7.7+...+2^{2008}.7\)
\(\Rightarrow A:7\)dư 1.
#Ngụy
#Fallen_Angel
Ta có : A = 1 + 2 + 22 + 23 + .... + 22009 + 22010
Đặt B = 2 + 22 + 23 + .... + 22009 + 22010
Khi đó A = 1 + B
Lại có : B = 2 + 22 + 23 + .... + 22009 + 22010
= (2 + 22 + 23) + (24 + 25 + 26) +.... + (22008 + + 22009 + 22010)
= (2 + 22 + 23) + 23.(2 + 22 + 23) + ... + 22007.(2 + 22 + 23)
= 14 + 23.14 + .... + 22007.14
= 14.(1 + 23 + ... + 22007)
= 2.7.(1 + 23 + ... + 22007) \(⋮7\)
=> \(B⋮7\)
=> (B + 1) : 7 dư 1
=> A : 7 dư 1
Vậy số dư khi A : 7 là 1
\(A=3^{12}+5^{13}+7^{15}+11^{2010}\)
\(=\left(3^4\right)^3+\left(...5\right)+\left(7^4\right)^3.7^3+\left(...1\right)\)
\(=\left(...1\right)^3+\left(...5\right)+\left(...1\right)^2.343+\left(...1\right)\)
\(=\left(...1\right)+\left(...5\right)+\left(...3\right)+\left(...1\right)\)
\(=\left(...0\right)\)$=\left(...0\right)$=(...0)chia 5 dư 0
Ta co : 312 tan cung la 1
513 tan cung la 5
715 tan cung la 1
112010tan cung la 1
Vay A co tan cung la :
5+1+1+1=8
Ma : 8:5=1 du 3
Vay A chia 5 du 3