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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: \(\frac{63}{10}>\frac{59}{9}hay6\frac{3}{10}>6\frac{5}{9}\)
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)
\(2+\frac{3}{10}+\frac{4}{100}\) \(23+\frac{5}{1000}\)
\(=\frac{200}{100}+\frac{30}{100}+\frac{4}{100}\) \(=\frac{23000}{1000}+\frac{5}{1000}\)
\(=\frac{234}{100}=2,34\) \(=\frac{23005}{1000}=23,005\)
\(11+\frac{7}{10}\) \(\frac{6}{10}+\frac{2}{100}\)
\(=\frac{110}{10}+\frac{7}{10}\) \(=\frac{60}{100}+\frac{2}{100}\)
\(=\frac{117}{10}=11,7\) \(=\frac{62}{100}=0,62\)
\(a,\frac{7}{13};\frac{7}{12};\frac{7}{10};\frac{7}{8}\)
\(b,\frac{9}{40};\frac{1}{4};\frac{3}{10};\frac{3}{8}\)
\(\frac{3}{10}+\frac{7}{100}=\frac{37}{100}\)
\(7+\frac{9}{1000}=\frac{7009}{1000}\)
\(\frac{3}{10}+\frac{7}{100}=\frac{30}{100}+\frac{7}{100}=\frac{37}{100}\)
\(7+\frac{9}{1000}=\frac{7000}{1000}+\frac{9}{1000}=\frac{7009}{1000}\)