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a) \(4\frac{3}{5}-\left(\frac{5}{2}-2\right)+\frac{5}{4}\)
\(=\frac{23}{5}-\frac{5}{2}+2+\frac{5}{4}\)
\(=\frac{107}{20}\)
b) \(47,31-18,27-8,27+4,6\)
\(=25,37\)
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
A = \(\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)
A = \(\frac{81+27+9+3+1}{81}\)
A = \(\frac{121}{81}\)
\(1,\)
\(a,1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(=\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)
\(=\frac{81+27+9+3+1}{81}\)
\(=\frac{121}{81}\)
\(=1\frac{40}{81}\)
\(b,37,25+4,7.2,3-9,8\)
\(=37,25+10,81-9,8\)
\(=38,26\)
\(2,\)
\(a,x:4,37=5,6\left(dư1,53\right)\)
\(x=5,6.4,37+1,53\)
\(\Rightarrow x=26,002\)
\(b,13,5.\left(x:5,6\right)=36,45\)
\(\left(x:5,6\right)=36,45:13,5\)
\(x:5,6=2,7\)
\(x=2,7.5,6\)
\(\Rightarrow x=15,12\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+.......+\frac{1}{59049}\)
\(3A=3.\left(\frac{1}{3}+\frac{1}{9}+......+\frac{1}{59049}\right)\)
\(3A=1+\frac{1}{3}+........+\frac{1}{19683}\)
\(3A-A=\left(1+\frac{1}{3}+......+\frac{1}{19683}\right)-\left(\frac{1}{3}+\frac{1}{9}+........+\frac{1}{59049}\right)\)
\(2A=1-\frac{1}{59049}\)
\(2A=\frac{59048}{59049}\)
\(A=\frac{59048}{59049}:2\)
\(A=\frac{59048}{118098}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+12,5\%:0,125-\frac{1}{10}:0,1=\frac{1}{2}:\frac{1}{2}-\frac{1}{4}:\frac{1}{4}+\frac{1}{8}:\frac{1}{8}-\frac{1}{10}:\frac{1}{10}=\frac{1}{2}.2-\frac{1}{4}.4+\frac{1}{8}.8-\frac{1}{10}.10=1-1+1-1=0\)
a) Cho: \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(\Rightarrow3A-A=3-\frac{1}{81}\)
\(\Rightarrow A=\frac{3-\frac{1}{81}}{2}\)
\(A=\frac{121}{81}\)
b) \(37,52+4,7\times2,3-9,8\)
\(=37,52+10,81-9,8\)
\(=38,53\)
Chúc bn học tốt !!!!!