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1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
1) \(4\frac{3}{10}=\frac{43}{10};21\frac{7}{100}=\frac{2107}{100};7\frac{39}{100}=\frac{739}{100};6\frac{123}{1000}=\frac{6123}{1000}\)
2)\(a,5\frac{2}{10}+7\frac{1}{10}=\frac{52}{10}+\frac{71}{10}=\frac{123}{10}\)
\(b,5\frac{6}{7}-3\frac{5}{7}=\frac{41}{7}-\frac{26}{7}=\frac{15}{7}\)
\(c,8\frac{3}{5}x2\frac{6}{7}=\frac{43}{5}x\frac{20}{7}=\frac{172}{7}\)
\(d,1\frac{3}{10}:5\frac{7}{8}=\frac{13}{10}:\frac{47}{8}=\frac{13}{10}x\frac{47}{8}=\frac{611}{80}\)
3) \(7\frac{9}{10}và4\frac{9}{10}\)
Ta có: \(7\frac{9}{10}=\frac{79}{10};4\frac{9}{10}=\frac{49}{10}\)
Suy ra: \(\frac{79}{10}>\frac{49}{10}hay7\frac{9}{10}>4\frac{9}{10}\)
\(6\frac{3}{10}và6\frac{5}{9}\)
Ta có: \(6\frac{3}{10}=\frac{63}{10};6\frac{5}{9}=\frac{59}{9}\)
Suy ra: \(\frac{63}{10}>\frac{59}{9}hay6\frac{3}{10}>6\frac{5}{9}\)