A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

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

\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)

Đặt P = ... ( biểu thức đề bài ) 

Nhận xét: Với \(k\inℕ^∗\) ta có: 

\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)

\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)

\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

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

\(=\frac{16}{35}\)

Ta có: 

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

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

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

    =\(11+\left[\left(\frac{5}{7}+\frac{2}{7}\right)+\left(\frac{3}{5}+\frac{1}{5}-\frac{4}{5}\right)\right]\)

    =\(11+\left[\frac{7}{7}+\frac{0}{5}\right]\)

    =\(11+\left[1+0\right]\)

    =\(11+1\)

     =12

ĐÃ GIẢI CHI TIẾT RÙI NHÉ. **** NHA

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)

ta có: A = \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)

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

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

             =     10                              +     \(\frac{106}{35}\)

             =           10                        +    \(3\frac{1}{35}\)

             =      \(13\frac{1}{35}\)

(2/5+2/7-2/11):(3/7-3/11+3/5) =2/5+2/7-2/11.7/3-11/3+5/3=2/1+2/1-2/1.1/3-1/3+1/3=2+1/3=7/3                                                                          Em đây mới học lớp 6 nên hay xem kĩ lại và tính bang máy tính  

a, 2/3

b,1

nhớ nhé