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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 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 \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
- \(\frac{3}{5}\)và \(\frac{5}{7}\)
Cách 1: Ta có : \(\frac{3}{5}=\frac{3x7}{5x7}=\frac{21}{35}\)
\(\frac{5}{7}=\frac{5x5}{7x5}=\frac{25}{35}\)
Vì \(21< 25\)\(\Rightarrow\frac{3}{5}< \frac{5}{7}\)
Cách 2: Ta có : \(\frac{3}{5}=1-\frac{2}{5}\)
\(\frac{5}{7}=1-\frac{2}{7}\)
Vì \(\frac{2}{5}>\frac{2}{7}\)\(\Rightarrow1-\frac{2}{5}< 1-\frac{2}{7}\)
hay \(\frac{3}{5}< \frac{5}{7}\)
Mấy phần kia cũng tương tự nhé . //Tại nó dài quá ấy,thông cảm nha//
_Học tốt_
b, \(\dfrac{7}{13}\) và \(\dfrac{17}{23}\)
\(\dfrac{7}{13}\) < \(\dfrac{7+10}{13+10}\) = \(\dfrac{17}{23}\)
Vậy \(\dfrac{7}{13}\) < \(\dfrac{17}{23}\)
\(\dfrac{12}{48}\) = \(\dfrac{12:12}{48:12}\) = \(\dfrac{1}{4}\)
\(\dfrac{13}{47}\) > \(\dfrac{13}{52}\) = \(\dfrac{13:13}{52:13}\) = \(\dfrac{1}{4}\)
Vậy \(\dfrac{12}{48}\) < \(\dfrac{13}{47}\)
Cách 1 : Quy đồng tử số :
\(\frac{2}{5}=\frac{4}{10}\) = \(\frac{4}{10}\) => \(\frac{2}{5}=\frac{4}{10}\)
Cách 2 : Nhân "tích chéo"
So sánh \(\frac{2}{5}\) với \(\frac{4}{10}\) nghĩa là so sánh 2 x 10 với 4 x 5.
Vì 2 x 10 = 20 = 4 x 5 nên \(\frac{2}{5}=\frac{4}{10}\)
Cách 3 : Rút gọn.
\(\frac{4}{10}=\frac{4:2}{10:2}=\frac{2}{5}\). Vậy \(\frac{2}{5}=\frac{4}{10}\)
3/8 <4/7
5/9<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 \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)