b, $B=10.(\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{35.37.39})$
$=10.\frac{1}{4}.(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{35.37}-\frac{1}{37.39})$
$=\frac{5}{2}.(\frac{1}{3}-\frac{1}{37.39})$
$=\frac{5}{2}.\frac{160}{481}$
$=\frac{400}{481}$.
a, Tử số $=\frac{1}{7}+\frac{1}{23}-\frac{1}{2009}$
$=\frac{23.2009+7.2009-23.7}{7.23.2009}$
Mẫu số $=\frac{23.2009+7.2009-23.7}{7.23.2009}+\frac{1}{7.23.2009}=\frac{23.2009+7.2009-23.7+1}{7.23.2009}$
Vậy $A=\frac{23.2009+7.2009-23.7}{7.23.2009}:\frac{23.2009+7.2009-23.7+1}{7.23.2009}+\frac{1}{30.2009-160}$
$=\frac{23.2009+7.2009-23.7}{7.23.2009}.\frac{7.23.2009}{23.2009+7.2009-23.7+1}+\frac{1}{30.2009-160}$
$=\frac{2009.(23+7)-23.7+1-1}{2009.(23+7)-23.7+1}+\frac{1}{30.2009-160}$
$=1-\frac{1}{30.2009-160}+\frac{1}{30.2009-160}$
$=1$.
