\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)

\(=\frac{14}{6}+\frac{1}{3}+\frac{14}{9}+\frac{7}{15}+\frac{14}{14}+\frac{6}{4}\)

\(=\frac{42+6+28}{18}+\frac{7}{15}+\frac{2}{2}+\frac{3}{2}\)

\(=\frac{76}{18}+\frac{7}{15}+\frac{5}{2}\)

\(=\frac{76}{18}+\frac{5}{2}+\frac{7}{15}\)

\(=\frac{76+45}{18}+\frac{7}{15}\)

\(=\frac{121}{18}+\frac{7}{15}\)

\(=\frac{605+42}{90}\)

\(=\frac{647}{90}\)

\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)

\(=(\frac{7}{3}+\frac{1}{3})+1+(\frac{14}{9}+\frac{7}{15})+\frac{3}{2}\)

\(=\frac{8}{3}+1+\frac{91}{45}+\frac{3}{2}\)

\(=2+\frac{2}{3}+1+2+\frac{1}{45}+1+\frac{1}{2}\)

\(=\left(\frac{2}{3}+\frac{1}{45}+\frac{1}{2}\right)+\left(2+1+2+1\right)\)

\(=1+\frac{17}{90}+6\)

\(=7+\frac{17}{90}\)

\(=7\frac{17}{90}\)

a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)

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

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

\(=1\)

b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)

\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)

\(=-\frac{5}{6}\)

Các câu khác làm tương tự thôi :))

\(\frac{1}{3}-|\frac{-1}{4}+\frac{5}{6}|-|\frac{-7}{12}|\)

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

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

\(=1\)

[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)

\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)

Ta có : 

\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)

\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(\Rightarrow P=\frac{3}{11}\)

Vậy \(P=\frac{3}{11}\)

\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)

đề bài của bn sai nên mk sửa luôn nha

a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)

                \(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}\)

                \(\frac{3}{x-1}=\frac{3}{7}\)

\(\Rightarrow x-1=7\)

              \(x=7+1\)

       Vậy \(x=8\)

b.\(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)

           \(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}\)

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

\(\Rightarrow2x+1=7\)

             \(2x=7-1\)

             \(2x=6\)

               \(x=6:2\)

\(x=3\)

a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)

             \(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}=\frac{3}{7}\)

           \(\Rightarrow x-1=7\)

                \(x=7+1=8\)

      VẬY, \(x=8\)

b. \(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)

            \(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}=\frac{7}{5}\)

          \(\Rightarrow2x+1=7\)

                        \(2x=7-1=6\)

                          \(x=6:2=3\)

            VẬY, \(x=3\)

a ) \(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)

\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{4}\)

\(\frac{3}{x-1}=\frac{6}{14}=\frac{3}{7}\)

\(=>x-1=7=>x=7+1=8\)

b ) \(\frac{42}{25}\div\frac{2x+1}{5}=\frac{6}{5}\)

\(\frac{2x+1}{5}=\frac{42}{25}\div\frac{6}{5}\)

\(\frac{2x+1}{5}=\frac{210}{150}=\frac{7}{5}\)

\(2x=7-1=6=>x=6\div2=3\)

Tính tất cả ra thì được:

\(=\frac{\frac{973}{60}}{\frac{139}{60}}:\frac{\frac{1255}{221}}{\frac{1506}{221}}+\frac{5858}{5050}\)

\(=\frac{\frac{139}{60}}{\frac{973}{60}}.\frac{\frac{1506}{221}}{\frac{1255}{221}}+\frac{5858}{5050}\)

Tính tử và mẫu dần rồi ra ( phần này dễ mà )

Ta được: ( mình chỉ lấy 2 chữ số phần thập phân thôi )

\(=\frac{1578}{9209}+\frac{5858}{5050}\)

 = 133/100

End 

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)

    \(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+\frac{2}{14.18}+...+\frac{2}{198.202}\)

     \(=\frac{1}{2}.\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+\frac{4}{14.18}+...+\frac{4}{198.202}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)

    \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

    \(=\frac{1}{2}.\frac{50}{101}=\frac{25}{101}\)

Ai đúng mình k

where ?

\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)

\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)

\(=\frac{7}{12}+\frac{31}{12}\)

\(=\frac{38}{12}=\frac{19}{6}\)

\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)

\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\frac{2}{13}\)

\(=\frac{-10}{117}\)

\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)

\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)

\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)

\(=\frac{4}{5}\cdot\frac{13}{7}\)

\(=\frac{52}{35}\)