tính nhanh nhé các bạn ! 

Tớ thấy bài này bạn nên nhóm các phân số có chung một dữ kiện nào đó với nhau rồi tính

Ta có: \(\frac{1}{x}-\frac{4}{6\cdot10}-\frac{3}{5\cdot6}-\frac{1}{16\cdot3}-\frac{1}{24\cdot7}-\frac{1}{28\cdot5}=\frac{1}{210}\)

=>\(\frac{1}{x}-\frac{1}{15}-\frac{1}{10}-\frac14\left(\frac{4}{12\cdot16}+\frac{4}{24\cdot28}\right)-\frac{1}{140}=\frac{1}{210}\)

=>\(\frac{1}{x}-\frac{2}{30}-\frac{3}{30}-\frac14\left(\frac{1}{12}-\frac{1}{16}+\frac{1}{24}-\frac{1}{28}\right)-\frac{1}{140}=\frac{1}{210}\)

=>\(\frac{1}{x}-\frac{5}{30}-\frac14\left(\frac{28}{336}-\frac{21}{336}+\frac{14}{336}-\frac{12}{336}\right)-\frac{1}{140}-\frac{1}{210}=0\)

=>\(\frac{1}{x}-\frac16-\frac14\cdot\frac{9}{336}-\frac{3}{420}-\frac{2}{420}=0\)

=>\(\frac{1}{x}-\frac16-\frac{1}{84}-\frac{3}{112\cdot4}=0\)

=>\(\frac{1}{x}-\frac{15}{84}-\frac{3}{448}=0\)

=>\(\frac{1}{x}-\frac{5}{28}-\frac{3}{448}=0\)

=>\(\frac{1}{x}-\frac{80}{448}-\frac{3}{448}=0\)

=>\(\frac{1}{x}=\frac{83}{448}\)

=>\(x=\frac{448}{83}\)

b) 

Gọi 3 số đó là : a) b) c)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)là số nguyên

Vì a ; b ; c số tự nhiên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)là phân số

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)lớn nhất \(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}< 2\)và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)nhỏ nhất \(>0\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Vậy 3 số tự nhiên cần tìm là : 2 ; 3 ; 6

a) 

\(A=\frac{4}{6}\times10+\frac{6}{10}\times16+\frac{1}{16}\times3+\frac{1}{24}\times7+\frac{1}{28}\times5\)

\(A=\frac{20}{3}+\frac{48}{5}+\frac{3}{16}+\frac{7}{24}+\frac{5}{28}\)

\(A=\frac{11200}{1680}+\frac{16128}{1680}+\frac{315}{1680}+\frac{490}{1680}+\frac{300}{1680}\)

\(A=\frac{26433}{1680}\)

Vậy \(A=\frac{26433}{1680}\)

\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)

\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)

\(c.\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{1}{4}\right):\frac{3}{7}\)

\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)

\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)

\(=\frac{7}{3}\times\left[\left(\frac{-3}{4}+\frac{1}{4}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)

\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)

\(=\frac{7}{3}\times\frac{1}{2}\)

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

\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)

\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)

\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)

\(=\frac{7}{8}\times\frac{44}{7}\)

\(=\frac{11}{2}\)