Question

Tính hợp lý:

a) A= \(\frac{101}{19}\). \(\frac{61}{218}\)\(-\) \(\frac{101}{218}\). \(\frac{42}{19}\)+ \(\frac{117}{218}\)

b) B= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\). \(\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)

Answer 1

a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)

        = \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)

        =\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)

        =\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)

        =\(\frac{101}{218}.1+\frac{117}{218}\)

        =\(\frac{101}{218}+\frac{117}{218}\)

        =\(\frac{218}{218}\)\(=1\)

b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)

        =     \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)

        = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)

        = \(0\)