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Ta có BCNN(6,8,9,3,2) = 72
Do đó \(\frac{5}{6}=\frac{5\cdot12}{6\cdot12}=\frac{60}{72}\)
\(\frac{13}{8}=\frac{13\cdot9}{8\cdot9}=\frac{117}{72}\)
\(\frac{8}{9}=\frac{8\cdot8}{9\cdot8}=\frac{64}{72}\)
\(\frac{1}{3}=\frac{1\cdot24}{3\cdot24}=\frac{24}{72}\)
\(\frac{3}{2}=\frac{3\cdot36}{2\cdot36}=\frac{108}{72}\)
=> \(\frac{117}{72}>\frac{108}{72}>\frac{64}{72}>\frac{60}{72}>\frac{24}{72}\)
hay \(\frac{13}{8}>\frac{3}{2}>\frac{8}{9}>\frac{5}{6}>\frac{1}{3}\)
Có \(\frac{5}{6}< 1;\frac{13}{8}>1;\frac{8}{9}< 1;\frac{1}{3}< 1;\frac{3}{2}>1\)
Do \(\frac{1}{3}=\frac{2}{6}< \frac{5}{6}\) nên \(\frac{1}{3}\) nhỏ nhất.
Ta cũng có \(\frac{5}{6}< \frac{5+3}{6+3}=\frac{8}{9}\) ( dựa lại vào BĐT \(\frac{a}{b}< \frac{a+c}{b+c}\) khi \(\frac{a}{b}< 1\)và \(a;b;c\ne0;\infty\))
Vì \(\frac{3}{2}=\frac{12}{8}< \frac{13}{18}\)nên \(\frac{13}{8}\)lớn nhất.
Đáp án là \(\frac{1}{3};\frac{5}{6};\frac{8}{9};\frac{3}{2};\frac{13}{8}\)
\(\dfrac{8}{9};\dfrac{7}{8};\dfrac{6}{7};\dfrac{5}{6};\dfrac{4}{5};\dfrac{3}{4};\dfrac{2}{3};\dfrac{1}{2}\)
a) Ta có: \(\dfrac{6}{11}=\dfrac{18}{33}\);
\(\dfrac{23}{33}=\dfrac{23}{33}\)
\(\dfrac{2}{3}=\dfrac{22}{33}\)
Do đó: \(\dfrac{6}{11}< \dfrac{2}{3}< \dfrac{23}{33}\)
b) Ta có: \(1>\dfrac{8}{9}>\dfrac{8}{11}\)
\(\dfrac{9}{8}=\dfrac{8}{8}>1\)
Do đó: \(\dfrac{9}{8}>\dfrac{8}{9}>\dfrac{8}{11}\)
a) \(\dfrac{6}{11};\dfrac{2}{3};\dfrac{23}{33}\)
b) \(\dfrac{9}{8};\dfrac{8}{9};\dfrac{8}{11}\)