a)A=(\(\frac13\) -\(\frac{8}{15}-\frac17\) )+(\(\frac23-\frac{-7}{15}\) \(+1\frac17\) )

=(\(\frac13\) -\(\frac{8}{15}-\frac17\) )+(\(\frac23-\frac{-7}{15}\) +\(\frac87)\)

=\(\frac13\) -\(\frac{8}{15}-\frac17\) +\(\frac23-\frac{-7}{15}\) +\(\frac87\)

=(\(\frac13\)+\(\frac23\) )+(\(\frac{8}{15}\) _\(\frac{-7}{15}\) )-(\(\frac87\) -\(\frac17\) )

=\(\frac33+\frac{15}{15}-\frac77\)

=1+1-1
=2-1 =1

Vậy A=1

b)B=0,25+\(\frac53\) ​−(\(\frac18\) ​−\(\frac25\) ​+\(1\frac14\) ​).

=\(\frac{25}{100}\) +\(\frac53\) ​−(\(\frac18\) ​−\(\frac25\) ​+\(\frac54\) ​)

=\(\frac{25}{100}\) +\(\frac53\) ​−(\(\frac{5}{40}-\frac{16}{40}+\frac{50}{40}\) ​)

=\(\frac{25}{100}\) +\(\frac53\) ​−\(\frac{39}{40}\)

=\(\frac14\) +\(\frac53\) ​−\(\frac{39}{40}\)

=\(\frac{23}{12}\) ​−\(\frac{39}{40}\)

=\(\frac{230}{120}-\frac{117}{120}\)

=\(\frac{113}{120}\)

Vậy B=\(\frac{113}{120}\)