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 \(A=\frac{26433}{1680}\)

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

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

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

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

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

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

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

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

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

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

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

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:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)

\(\Rightarrow A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)

\(\Rightarrow A=\left(\frac{1}{6}+\frac{1}{10}\right)+\left(\frac{1}{7}+\frac{1}{9}\right)+\frac{1}{8}\)

\(\Rightarrow A=\left(\frac{10}{6.10}+\frac{6}{6.10}\right)+\left(\frac{9}{7.9}+\frac{7}{7.9}\right)+\frac{8}{8.8}\)

\(\Rightarrow A=\frac{16}{6.10}+\frac{16}{7.9}+\frac{8}{8.8}\)

\(\Rightarrow A=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)\)

Ta lại có:

\(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)

\(\Rightarrow B=\left(\frac{1}{6.10}+\frac{1}{6.10}\right)+\left(\frac{1}{7.9}+\frac{1}{7.9}\right)+\frac{1}{8.8}\)

\(\Rightarrow B=\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\)

Vậy : 

\(A:B=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right):\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)=8\)

Vậy \(A:B=8\)