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\(A=4^{17}+4^{18}+4^{19}+4^{20}+4^{17}\left(999-4\right)\)
\(=4^{17}+4^{18}+4^{19}+4^{20}+999.4^{17}-4^{18}\)
\(=4^{17}+4^{19}+4^{20}+999.4^{17}\)
\(=4^{17}\left(1+4^2+4^3\right)+999.4^{17}\)
\(=81.4^{17}+999.4^{17}\)
\(\left\{{}\begin{matrix}81⋮9\\999⋮9\end{matrix}\right.\Rightarrow A⋮9\)
\(8^{30}+8^{31}+8^{32}\)
\(=8^{30}.1+8^{30}.8+8^{30}.8^2\)
\(=8^{30}.1+8^{30}.8+8^{30}.64\)
\(=8^{30}\left(1+8+64\right)\)
\(=8^{30}.73\)
\(=\left(2^3\right)^{30}.73\)
\(=2^{90}.73\)
\(=2^{89}.146⋮146\rightarrowđpcm\)
\(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)
\(=4^{25}.1+4^{25}.4+4^{25}.4^2+4^{25}.4^3+4^{25}.4^4+4^{25}.4^5\)
\(=4^{25}.1+4^{25}.4+4^{25}.16+4^{25}.64+4^{25}.256+4^{25}.1024\)
\(=4^{25}\left(1+4+16+64+256+1024\right)\)
\(=4^{25}.1365\)
\(=4^{25}.195.7⋮7\rightarrowđpcm\)
b)\(B=\dfrac{3}{2}+\dfrac{13}{12}+\dfrac{31}{30}+...+\dfrac{9901}{9900}\)
\(=1+\dfrac{1}{2}+1+\dfrac{1}{12}+1+\dfrac{1}{30}+...+1+\dfrac{1}{9900}\)
\(=1+1+1+...+1\left(50cs\right)+\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(=50+\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)
\(C=\dfrac{5}{6}+\dfrac{19}{20}+\dfrac{41}{42}+...+\dfrac{10099}{10100}\)
\(=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{42}\right)+...+\left(1-\dfrac{1}{10100}\right)\)
\(=1+1+...+1\left(50cs\right)-\dfrac{1}{6}-\dfrac{1}{20}-\dfrac{1}{42}-...-\dfrac{1}{10100}\)
\(B-C=\left(50+\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)-\left(50-\dfrac{1}{6}-\dfrac{1}{20}-...-\dfrac{1}{10100}\right)\)
\(=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}+\dfrac{1}{10100}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)
Chúc Bạn Học Tốt và Đạt nhiều thành tích tốt !!!
Bài 2:
a) \(9^{1945}-2^{1930}\)
Ta có:
\(\left\{{}\begin{matrix}9^{1945}=\left(9^5\right)^{389}=\overline{.......9}\\2^{1930}=\left(2^{10}\right)^{193}=\overline{.......4}\end{matrix}\right.\)
\(\Rightarrow\overline{........9}-\overline{.........4}=\overline{..........5}.\)
Vì \(\overline{.......5}⋮5\) nên \(\overline{.........9}-\overline{........4}=\overline{........5}\)
\(\Rightarrow9^{1945}-2^{1930}⋮5\left(đpcm\right).\)
Chúc bạn học tốt!
Bài 1 : \(3^{n+2}\)\(-2^{n+2}\)+ \(3^n-2^n\)= \(\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
= \(3^n\)\(\left(3^2+1\right)\) \(-2^n\left(2^2+1\right)\)= \(3^n\times10-2^{n-1}\times10\)
= 10 \(\times\left(3^n+2^{n+1}\right)\)
chia hết cho 10
Bài 2 :
\(A=75.\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\) =\(75+25+75.4.\left(4^{2003}+4^{2003}+....+4^2+4\right)\)
= \(100+300.\left(4^{2003}+4^{2003}+...+4^2+4\right)\)
chia het cho 100