\(\cos87^020'=\sin2^040'< \sin40^0< \sin42^0< \sin65^025'=\cos24^035'< \sin78^0\)

TA có: \(\frac{3}{400}=\frac{3\times56}{400\times56}=\frac{168}{22400}\)

\(\frac{-6}{217}=\frac{6}{-217}=\frac{6\cdot28}{-217\cdot28}=\frac{168}{-6076}\)

\(\frac{-7}{-284}=\frac{7}{284}=\frac{7\cdot24}{284\cdot24}=\frac{168}{6816}\)

\(\frac{8}{-305}=\frac{8\cdot21}{-305\cdot21}=\frac{168}{-6405}\)

Ta có: 6405<6076

=>\(\frac{168}{6405}>\frac{168}{6076}\)

=>\(-\frac{168}{6405}<-\frac{168}{6076}\)

=>\(\frac{8}{-305}<\frac{-6}{217}<0\left(1\right)\)

Ta có: 22400>6816

=>\(\frac{168}{22400}<\frac{168}{6816}\)

=>\(0<\frac{3}{400}<\frac{-7}{-284}\) (2)

Từ (1),(2) suy ra \(\frac{8}{-305}<\frac{-6}{217}<\frac{3}{400}<\frac{-7}{-284}\)

\(\dfrac{12}{11};\dfrac{6}{5};\dfrac{5}{4};\dfrac{4}{3};\dfrac{3}{2}\)

\(\dfrac{3}{5}=\dfrac{36}{60};\dfrac{3}{4}=\dfrac{45}{60};\dfrac{7}{12}=\dfrac{35}{60}\\ \dfrac{35}{60}< \dfrac{36}{60}< \dfrac{45}{60}\Rightarrow\dfrac{7}{12}< \dfrac{3}{5}< \dfrac{3}{4}\)

Sắp xếp theo thứ tự từ bé đến lớn 

9/5;1/6;15/7

Sắp xếp các phân số sau theo thứ tự tăng dần 12 / 7, 2/3 9/4

\(\frac{2}{3};\frac{12}{7};\frac{9}{4}\)

HT