Question

(\(\frac{1}{21}\)+\(\frac{1}{210}\)+\(\frac{1}{2010}\))*(\(\frac{1}{3}\)-\(\frac{1}{30}\)-\(\frac{1}{5}\)-\(\frac{1}{10}\))

Answer 1

Ta có:

     \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)

=   \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)

=   \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10-1-6-3}{30}\right)\)

=   \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(0\)

=    \(0\)

Answer 2

ta có \(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}=\frac{10-1-6-3}{30}=\frac{0}{30}=0\)

=>\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)x\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)=0\)

Answer 3

\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\times\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)

\(=\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\times\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)

\(=\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\times\left(\frac{10-1-6-3}{30}\right)\)

\(=\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\times\frac{0}{30}\)

\(=\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\times0\)

\(=0\)