\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}.\)

\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

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

\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)

\(\Leftrightarrow x\cdot\frac{5}{3}=15\)

\(\Leftrightarrow x=15:\frac{5}{3}\)

\(\Leftrightarrow x=15\cdot\frac{3}{5}\)

\(\Leftrightarrow x=9.\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)

\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)

\(\Rightarrow x.\frac{5}{3}=14+1=15\)

\(\Rightarrow x=15:\frac{5}{3}=9\)

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b) \(\left(x\cdot\frac{5}{3}-1\right):9=3\frac{1}{2}:2,25\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{7}{2}:\frac{9}{4}\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{14}{9}\)

\(x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

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

\(x\cdot\frac{5}{3}=14+1\)

\(x\cdot\frac{5}{3}=15\)

\(x=15:\frac{5}{3}\)

\(x=9\)

Vậy \(x=9\)

Đề còn thiếu 1 điều kiện nữa là \(n>0\)

Đặt \(A=\frac{4}{5.2!}+\frac{4}{5.3!}+\frac{4}{5.4!}+...+\frac{4}{5.n!}\) ta có : 

\(A=\frac{4}{5}\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\right)\)

Để \(A< 0,8\) thì \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}< 1\)

Đặt \(B=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\) ta có : 

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}+\frac{1}{n}\)

\(B< 1-\frac{1}{n}< 1\)

\(\Rightarrow\)\(B< 1\) ( đpcm ) 

Suy ra : \(A=\frac{4}{5}.B=0,8.B< 0,8\) ( vì \(B< 1\) ) 

Vậy \(\frac{4}{5.2!}+\frac{4}{5.3!}+\frac{4}{5.4!}+...+\frac{4}{5.n!}< 0,8\)

Chúc bạn học tốt ~ 

\(\frac{1}{3}+x\times\frac{2}{7}=\frac{11}{12}\)

\(x\times\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)

\(x\times\frac{2}{7}=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac{2}{7}\)

\(x=\frac{49}{24}\)

~Moon~

\(\frac{1}{3}+x.\frac{2}{7}=\frac{11}{12}\)

\(x.\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)

\(x.\frac{2}{7}=\frac{7}{12}\)

\(x=\frac{7}{12}\div\frac{2}{7}\)

\(x=\frac{49}{24}\)

bạn làm thế này nhé

 1/3 + x . 2/7 = 11/12

<=> 2x/7=11/12 - 1/3

<=> 2x/7= 7/12

<=> 2x.12=7.7

<=> 24x = 49

<=> x = 49/24

    Vậy x = 49/24

5.(1/5+1/17)-(2/5+2/17+9/15+12/68)

=5.22/85-22/17

=22/17-22/17

=0

Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)

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

\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)

\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)

\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)

\(=\)\(0\)

     Vậy ... 

                  Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^

Ai giúp mình tặng 10 hứa

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

\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{4}{5}+\frac{4}{5}\)

\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{8}{5}=\frac{18}{5}\)

\(\Rightarrow\frac{6}{5}y=\frac{18}{5}-\frac{4}{3}\)

\(\Rightarrow\frac{6}{5}y=\frac{34}{15}\)

\(\Rightarrow y=\frac{34}{15}:\frac{6}{5}\)

\(\Rightarrow y=\frac{34}{15}.\frac{5}{6}=\frac{17}{9}\)

=4/3 + 6/5 . y - 4/5 =14/5

 4/3 +6/5 . y           =14/5+4/5

4/3 +6/5 . y            =18/5

                 y           =18/5 : (4/3 + 6/5)

                 y           =18/5 :  38/15

                 y            = 9/19

    CHÚC BẠN HỌC TỐT !

< 1 nhé 

Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)

Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)

=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)

=> A < 1