a, 2016.(-4) - (-1008).8= (-1008).8 - (-1008).8=0

b, Sai đề à bạn

\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)

\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)

\(=17.\frac{1}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)

\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(=\frac{17}{3}.\frac{6}{49}\)

\(=\frac{34}{49}\)

\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)

\(=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)

\(=13.\frac{1}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)

\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(=\frac{13}{3}.\frac{6}{49}\)

\(=\frac{26}{49}\)

Câu C sai đề rồi , phải là như thế này :

\(C=\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)

\(=\frac{4}{5}.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)

\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)

\(=\frac{4}{5}.\left(\frac{1}{8}-\frac{1}{258}\right)\)

\(=\frac{4}{5}.\frac{125}{1032}\)

\(=\frac{25}{258}\)

\(\dfrac{\Leftrightarrow3A}{17}=\left(\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}+\dfrac{30}{49.79}+\dfrac{47}{79.126}\right)x=\dfrac{289}{9}\)

chết ấn nhầm nút

\(\dfrac{3}{17}A=\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}\)

\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\)

\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)

\(A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{6.17}{49.3}=\dfrac{34}{49}\)