a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)

\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)

\(A=\dfrac{-1}{10.2}\)

\(A=\dfrac{-1}{20}\)

b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)

\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)

\(B=\dfrac{2}{3.3}\)

\(B=\dfrac{2}{9}\)

c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)

\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)

\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)

\(C=\dfrac{2}{3}+\dfrac{2}{3}\)

\(C=\dfrac{4}{3}\)

d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)

\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)

\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)

\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)

\(D=\dfrac{-57}{57}=-1\)

e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)

\(E=\dfrac{27}{182}-\dfrac{27}{182}\)

\(E=0\)

a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)

\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)

\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)

\(=1+1\)

\(=2\)

b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)

\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)

\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)

\(=3+2+2\)

\(=7\)

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(=1-\dfrac{1}{7}\)

\(=\dfrac{6}{7}\)

a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\) (dấu chấm là dấu nhân)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

b) \(\frac{19}{13}+\frac{14}{6}+\frac{1}{9}+\frac{4}{6}+\frac{7}{13}+\frac{17}{9}\)

\(=\left(\frac{19}{13}+\frac{7}{13}\right)+\left(\frac{14}{6}+\frac{4}{6}\right)+\left(\frac{1}{9}+\frac{17}{9}\right)\)

= 2 + 3 + 2

= 7

= 1/1x2+1/2x3+...+1/7x8

= 1-1/2+1/2-...-1/8

= 1-1/8

=7/8

a/ \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}=\)

= 1-1/8 = 7/8

b/ = (17/9+1/9)+(19/13+7/13)+(14/6+10/6) = 18/9 + 26/13 + 24/6 = 2+2+4 = 8

\(a)\) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\)\(1-\frac{1}{8}\)

\(=\)\(\frac{7}{8}\)

Áp dụng công thức \(\frac{a}{b\left(b+a\right)}=\frac{1}{b}-\frac{1}{b+a}\) \(\left(a,b>0\right)\) nhé 

Chúc bạn học tốt ~ 

lam giup minh voi

Nếu biết cách đăng hình vào câu trả lời thì mk sẽ đăng cách giải và đáp số, chứ mk ngại viết lắm

ban giai ho minh voi

1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+9+8+9+10+12+13+14+15+16+17+18+19+20+30+1000000=10000274

10000274

10000274 nha

c= - 1/10 nha