\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Leftrightarrow x=2\)

\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

Đặt \(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

\(\Leftrightarrow\frac{5}{14}.\frac{4}{3}x=\frac{5}{14}\)

\(\Rightarrow\frac{4}{3}.x=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow x=2\)

Đúng rùi bn

BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6

\(\frac{1}{3}-\left(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{7.8}\right)=x-\frac{5}{18}\)

\(x-\frac{5}{18}=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(x-\frac{5}{18}=\frac{1}{3}-\frac{1}{3}+\frac{1}{8}\)

\(x-\frac{5}{18}=0+\frac{1}{8}\)

\(x-\frac{5}{18}=\frac{1}{8}\)

\(x=\frac{1}{8}+\frac{5}{18}\)

\(x=\frac{9}{72}+\frac{20}{72}\)

\(x=\frac{29}{72}\)

1/3 - 1/12 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18

            1/4 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18

                        1/5 - 1/30 - 1/42 - 1/56 = x - 5/18

                                    1/6 - 1/42 - 1/56 = x - 5/18

                                                1/7 - 1/56 = x - 5/18

                                                            1/8 = x - 5/18

                                                                x=1/8+5/18

                                                                x= 29/72

                                                           Vậy : x = 29/72

\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)

\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)

\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{11}{6}\)

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

 \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)

\(x+A=\frac{47}{42}\)

     ta thấy :

              \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

              \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

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

               \(A=\frac{1}{1}-\frac{1}{6}\)

             \(A=\frac{5}{6}\)

       vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)       

  hay  \(x+\frac{5}{6}=\frac{47}{42}\)

     \(x=\frac{47}{42}-\frac{5}{6}\)

 \(x=\frac{2}{7}\)

\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)

\(x=\frac{47}{42}-\frac{5}{6}\)

\(x=\frac{2}{7}.\)

\(\left(x+50\%\right):\frac{7}{8}=\frac{5}{7}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)=\frac{5}{7}.\frac{7}{8}\)

\(\Rightarrow x+\frac{1}{2}=\frac{5}{8}\)

\(\Rightarrow x=\frac{5}{8}-\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{8}\)
Vậy...

Mình làm tiếp bài của bạn " I have a crazy idea "

b) \(\frac{25-x}{3}=\frac{15}{2}\)

Áp dụng tỉ lệ thức:

\(\left(25-x\right).2=15.3\)

\(\Rightarrow25-x=\frac{15.3}{2}=\frac{45}{2}\Leftrightarrow x=25-\frac{45}{2}=\frac{5}{2}\)

c) \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)

\(\Rightarrow x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{7}\right)=1\Leftrightarrow x-\frac{6}{7}=1\Leftrightarrow x=1+\frac{6}{7}=\frac{13}{7}\)

c) x=-2 nha

d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)

=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)

=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)

bạn ơi kết quả là 7/60

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}\)

\(=\frac{7}{60}\)