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a; Cách một:
\(\dfrac{2}{9}\) = \(\dfrac{2\times2}{9\times2}\) = \(\dfrac{4}{18}\) < \(\dfrac{4}{10}\) Vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times3}{9\times3}\) = \(\dfrac{12}{27}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times2}{10\times2}\) = \(\dfrac{12}{20}\)
Vì \(\dfrac{12}{27}\) < \(\dfrac{12}{20}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{12}{20}\)
\(\dfrac{3}{8}\) = \(\dfrac{3\times4}{8\times4}\) = \(\dfrac{12}{24}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times3}{7\times3}\) = \(\dfrac{12}{21}\)
Vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
\(\dfrac{5}{9}\) = \(\dfrac{5\times7}{9\times7}\) = \(\dfrac{35}{63}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times5}{10\times5}\) = \(\dfrac{35}{50}\)
Vì \(\dfrac{35}{63}\) < \(\dfrac{35}{50}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách hai:
a; \(\dfrac{2}{9}\) = \(\dfrac{2\times10}{9\times10}\) = \(\dfrac{20}{90}\); \(\dfrac{4}{10}\) = \(\dfrac{4\times9}{10\times9}\) = \(\dfrac{36}{90}\)
Vì \(\dfrac{20}{90}\) < \(\dfrac{36}{90}\) vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
b; \(\dfrac{4}{9}\) = \(\dfrac{4\times10}{9\times10}\) = \(\dfrac{40}{90}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times9}{10\times9}\) = \(\dfrac{54}{90}\)
Vì \(\dfrac{40}{90}\) < \(\dfrac{54}{90}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{6}{10}\)
c; \(\dfrac{3}{8}\) = \(\dfrac{3\times7}{8\times7}\) = \(\dfrac{21}{56}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times8}{7\times8}\) = \(\dfrac{32}{56}\)
Vì \(\dfrac{21}{56}\) < \(\dfrac{32}{56}\) vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
d; \(\dfrac{5}{9}\) = \(\dfrac{5\times10}{9\times10}\) = \(\dfrac{50}{90}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times9}{10\times9}\) = \(\dfrac{63}{90}\)
Vì \(\dfrac{50}{90}\) < \(\dfrac{63}{90}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách 1:
\(\dfrac{3}{8}\) = \(\dfrac{3\times4}{8\times4}\) = \(\dfrac{12}{32}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times3}{7\times3}\) = \(\dfrac{12}{21}\)
Vì \(\dfrac{12}{32}\) < \(\dfrac{12}{21}\) nên \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
\(\dfrac{5}{9}\) = \(\dfrac{5\times7}{9\times7}\) = \(\dfrac{35}{63}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times5}{5\times10}\) = \(\dfrac{35}{50}\)
Vì \(\dfrac{35}{65}\) < \(\dfrac{35}{50}\) nên \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách một:
\(\dfrac{4}{10}\) = \(\dfrac{4:2}{10:2}\) = \(\dfrac{2}{5}\) > \(\dfrac{2}{9}\) Vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times3}{9\times3}\) = \(\dfrac{12}{27}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times2}{10\times2}\) = \(\dfrac{12}{20}\)
Vì \(\dfrac{12}{27}\) < \(\dfrac{12}{20}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{12}{20}\)
Cách hai:
\(\dfrac{2}{9}\) = \(\dfrac{2\times10}{9\times10}\) = \(\dfrac{20}{90}\); \(\dfrac{4}{10}\) = \(\dfrac{4\times9}{10\times9}\) = \(\dfrac{36}{90}\)
Vì \(\dfrac{20}{90}\) < \(\dfrac{36}{90}\) vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times10}{9\times10}\) = \(\dfrac{40}{90}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times9}{10\times9}\) = \(\dfrac{54}{90}\)
Vì \(\dfrac{40}{90}\) < \(\dfrac{54}{90}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{6}{10}\)
a; Cách một:
\(\dfrac{2}{9}\) = \(\dfrac{2\times2}{9\times2}\) = \(\dfrac{4}{18}\) < \(\dfrac{4}{10}\) Vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times3}{9\times3}\) = \(\dfrac{12}{27}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times2}{10\times2}\) = \(\dfrac{12}{20}\)
Vì \(\dfrac{12}{27}\) < \(\dfrac{12}{20}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{12}{20}\)
\(\dfrac{3}{8}\) = \(\dfrac{3\times4}{8\times4}\) = \(\dfrac{12}{24}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times3}{7\times3}\) = \(\dfrac{12}{21}\)
Vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
\(\dfrac{5}{9}\) = \(\dfrac{5\times7}{9\times7}\) = \(\dfrac{35}{63}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times5}{10\times5}\) = \(\dfrac{35}{50}\)
Vì \(\dfrac{35}{63}\) < \(\dfrac{35}{50}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách hai:
a; \(\dfrac{2}{9}\) = \(\dfrac{2\times10}{9\times10}\) = \(\dfrac{20}{90}\); \(\dfrac{4}{10}\) = \(\dfrac{4\times9}{10\times9}\) = \(\dfrac{36}{90}\)
Vì \(\dfrac{20}{90}\) < \(\dfrac{36}{90}\) vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
b; \(\dfrac{4}{9}\) = \(\dfrac{4\times10}{9\times10}\) = \(\dfrac{40}{90}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times9}{10\times9}\) = \(\dfrac{54}{90}\)
Vì \(\dfrac{40}{90}\) < \(\dfrac{54}{90}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{6}{10}\)
c; \(\dfrac{3}{8}\) = \(\dfrac{3\times7}{8\times7}\) = \(\dfrac{21}{56}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times8}{7\times8}\) = \(\dfrac{32}{56}\)
Vì \(\dfrac{21}{56}\) < \(\dfrac{32}{56}\) vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
d; \(\dfrac{5}{9}\) = \(\dfrac{5\times10}{9\times10}\) = \(\dfrac{50}{90}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times9}{10\times9}\) = \(\dfrac{63}{90}\)
Vì \(\dfrac{50}{90}\) < \(\dfrac{63}{90}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Câu 6:
Giải:
\(\dfrac{48}{64}\) = \(\dfrac{48:16}{64:16}\) = \(\dfrac{3}{4}\)
Vì 100 : 3 = 33 dư 1
Số lớn nhất có hai chữ số chia hết cho 3 là: 100 - 1 = 99
99 : 3 = 33
Vậy có 33 phân số có tử số nhỏ hơn 100 và có giá trị bằng \(\dfrac{48}{64}\)
Do \(\dfrac{48}{64}\) là phân số trùng với \(\dfrac{48}{64}\) nên thực tế số phân số có tử số nhỏ hơn 100 và có giá trị bằng \(\dfrac{48}{64}\) là:
33 - 1 = 32 (phân số)
Đáp số: 32 phân số
a) 1/2
b) 18/17; 17/16; 16/15; 15/14; 14/13; 13/12; 10/9
Nhớ k cho mk nha
a)1/2
b)Vì 18/17 <17/16 <16/15<15/14< 14/13< 13/12< 10/9
Nên, ta có:
18/17; 17/16; 16/15; 15/14; 14/13; 13/12; 10/9.
1a)\(\frac{5}{3}\)=\(\frac{5x4}{3x4}\)=\(\frac{20}{12}\); \(\frac{1}{4}\)=\(\frac{1x3}{4x3}\)=\(\frac{3}{12}\)
b)\(\frac{3}{8}\)=\(\frac{3x3}{8x3}\)=\(\frac{9}{24}\); \(\frac{7}{24}\)
c)\(\frac{1}{2}\)=\(\frac{1x15}{2x15}\)=\(\frac{15}{30}\); \(\frac{2}{3}\)=\(\frac{2x10}{3x10}\)=\(\frac{20}{30}\); \(\frac{3}{5}\)=\(\frac{3x6}{5x6}\)=\(\frac{18}{30}\)
2a)\(\frac{11}{8}\)>\(\frac{11}{9}\)
b)\(\frac{4}{9}\)<\(\frac{3}{5}\)
c)\(\frac{6}{5}\)>\(\frac{5}{6}\)