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\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\frac{2.2}{3}=\frac{4}{3}\)
câu sau tương tự nhé
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
\(C=\frac{11\cdot12+22\cdot24+44\cdot48}{33\cdot36+66\cdot72+132\cdot144}=\frac{12\left(11+44+132\right)}{12\left(99+396+1584\right)}=\frac{187}{2079}\)
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
\(A=\frac{3^7\cdot17-3^9}{2^3\cdot3^5}=\frac{3^7\left(17-3^2\right)}{2^3\cdot3^5}=\frac{3^7\cdot2^3}{2^3\cdot3^5}=9\)
\(B=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{3^2\cdot2^{36}}{2^{35}\cdot11-2^{36}}=\frac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2\cdot2^{36}}{2^{35}\cdot3^2}=2\)
\(\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2\cdot3^{28}}=\frac{3^{29}\cdot8}{2^2\cdot3^{28}}=6\)
c) \(\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{11\cdot2^{35}-2^{26}}\)
\(=\frac{9\cdot2^4\cdot2^{32}2^{ }}{\left(11-2\right)\cdot2^{35}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{9\cdot2^{35}}\)
\(=\frac{9\cdot1\cdot2^{32}}{9\cdot2^{31}}=\frac{2^{32}}{2^{31}}=2\)
