1, cho C=\(\frac{9}{2.5}+\frac{9}{5.8}+...+\frac{9}{29+32}\). C/m C<1
Giải:
C=3 . ( \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29+32}\))
C= 3 . (\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\))
C= 3 . (\(\frac{1}{2}-\frac{1}{32}\))
C= 3 . \(\frac{15}{32}\)
C= \(\frac{45}{32}\)>1
Vậy C > 1 (đpcm)