\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(1+3y=-4\)

\(\Leftrightarrow\)\(3y=-5\)

\(\Leftrightarrow\)\(y=-\frac{5}{3}\)

Vậy...

Ta có : 

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y.3=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.3=-4\)

\(\Rightarrow\left(\frac{5}{120}-\frac{4}{120}\right).120+y.3=-4\)

\(\Rightarrow\frac{1}{120}.120+y.3=-4\)

\(\Rightarrow1+y.3=-4\)

\(\Rightarrow3y=-4-1\)

\(\Rightarrow3y=-5\)

\(\Rightarrow y=-\frac{5}{3}\)

Vậy \(y=-\frac{5}{3}\)

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4.\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.\frac{3}{1}=-4\)

\(\Leftrightarrow\frac{1}{120}.120+3y=-4\)

\(\Leftrightarrow1+3y=-4\Leftrightarrow3y=-5\)

\(\Leftrightarrow y=\frac{-5}{3}\)

a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow1+x:\frac{1}{3}=-4\)

\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)

\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)

b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)

\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)

\(\Rightarrow x=4\)

y=-5/3 

y=-5/3

y=-5/3

 Câu a) :

x=-5/3

Câu b) :

GỢI Ý : 3n-5 phải chia hết cho n-4 để A là số nguyên ( đk : n khác 4)

\(a,\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+3x=-4\)

\(\left(\frac{1}{24}-\frac{1}{30}\right).120+3x=-4\)

\(\frac{1}{120}.120+3x=-4\)

\(1+3x=-4\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=-\frac{5}{3}\)

\(b,A=\frac{3n-5}{n-4}=\frac{3n-12+7}{n-4}=3+\frac{7}{n-4}\)

Để \(A\in Z\Rightarrow7⋮n-4\Leftrightarrow n-4\in\left(1;-1;7;-7\right)\)

\(\Rightarrow n\in\left(5;3;11;-3\right)\)

           \(\)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)  \(\)

\(\Rightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(1+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(3x=-3\)

\(\Rightarrow\)\(x=-1\)

\(\frac{1}{24\cdot25}+\frac{1}{25\cdot26}+...+\frac{1}{29\cdot30}\)

\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)

\(=\frac{1}{24}-\frac{1}{30}\)

\(=\frac{1}{120.}\)

\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)

\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)

\(=\frac{1}{24}-\frac{1}{30}\)

\(=\frac{1}{120}\)

Ủng hộ mk nha ^_-

\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)

\(=\frac{1}{24}-\frac{1}{30}=\frac{1}{120}\)

Lời giải:

\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}=\frac{25-24}{24.25}+\frac{26-25}{25.26}+...+\frac{30-29}{29.30}\)

\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)

\(=\frac{1}{24}-\frac{1}{30}=\frac{1}{120}\)

Vậy:

\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(1+x:\frac{1}{3}=-4\)

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

\(x=-15\)

\(\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right).120+x:\dfrac{1}{3}=-4\) 

\(\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\) 

                              \(\left(\dfrac{1}{24}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\) 

                                            \(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\) 

                                                        \(1+x:\dfrac{1}{3}=-4\) 

                                                               \(x:\dfrac{1}{3}=-4-1\) 

                                                               \(x:\dfrac{1}{3}=-5\) 

                                                                     \(x=-5.\dfrac{1}{3}\) 

                                                                     \(x=\dfrac{-5}{3}\)