Question

giúp mình với:<

Answer 1

\(a,A=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{49}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{49}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{49}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{49}\right)}=\dfrac{1}{4}\\ b,A=\dfrac{-\dfrac{6}{7}+\dfrac{6}{19}-\dfrac{6}{31}}{\dfrac{9}{7}-\dfrac{9}{19}+\dfrac{9}{31}}=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{19}+\dfrac{1}{31}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{19}+\dfrac{1}{31}\right)}=\dfrac{-6}{9}=-\dfrac{2}{3}\)

Answer 2