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20112-(304+2012)+(2013+304)
=20112-304-2012+2013+304
=20112+(-2012+2013)+(-304+304)
=20112+1+0=20113
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.25^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(25^2\right)^3.\left(2^3\right)^3.3^3}=\)\(\frac{3^{28}.25^5.2^{21}}{2^{12}.2^9.3^{24}.3^3.25^6}=\frac{3^{28}.25^5.2^{21}}{2^{21}.3^{27}.25^6}\)\(=\frac{3}{25}\)
a) 2012 - ( 304 + 2012 ) + ( 2013 + 304 )
= 2012 - 304 - 2012 + 2013 + 304
= 2012 + ( - 304 ) + ( - 2012 ) + 2013 + 304
= [ 2012 + ( - 2012 ) ] + [ ( - 304 ) + 304 ] + 2013
= 0 + 0 + 2013
= 2013
b) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)
\(=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{\left(3^2.2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{3^{24}.2^{12}.5^{12}.2^9.3^3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{3^{27}.5^{12}.2^{21}}\)
\(=\frac{3}{5^2}=\frac{3}{25}\)
a) 2012 - ( 304 + 2012 ) + (2013 + 304 )
= 2012 - 304 +2012 + 2013 + 304
= ( 2012 - 2012 ) + ( 304 + 304 ) + 2013
= 0 + 608 + 2013
= 2621
Chờ một chút để minh suy nghĩ
Xét N có:
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta các số hạng của M và N có:
\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)
Từ (1);(2);(3) => M > N
trong câu hỏi tương tự có một bài giốngđè và được giải rồi, bạn xem thử đi
\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{\left(3^2\right)^{14}\cdot\left(5^2\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(3\cdot2^3\right)^3}\)
\(=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot3^3\cdot2^9}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{25}\cdot5^{12}\cdot2^{21}}=\frac{3^3}{5^2}=\frac{27}{25}\)
\(\frac{9^{14}}{18^{12}}.\frac{25^5}{625^3}.\frac{8^7}{24^3}\)
\(=\frac{9^{14}}{\left(9.2\right)^{12}}.\frac{25^5}{25^6}.\frac{8^7}{\left(8.3\right)^3}\)
\(=\frac{9^{14}}{9^{12}.2^{12}}.\frac{1}{25}.\frac{8^7}{8^3.3^3}\)
\(=\frac{9^2}{2^{12}}.\frac{1}{25}.\frac{8^4}{3^3}\)
\(=\frac{81}{4096}.\frac{1}{25}.\frac{4096}{27}\)
\(=\frac{81}{4096}.\frac{4096}{27}.\frac{1}{24}=3.\frac{1}{24}=\frac{3}{24}\)
**** **** ****
2012-304-2012+2013+304
=(2012-2012)+(304-304)+2013
=0+0+2013
=2013
2012 - ( 304 + 2012 ) + ( 2013 + 304 )
= 2012 - 2316 + 2317
= -304 + 2317
= 2013