\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2020-2019}{2019.2020}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)

mik làm luôn nhé

=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2019-1/2020

=(-1/3+1/3)+(-1/4+1/4)+...+(-1/2019+1/2019)+(1/2-1/2020)

=0+0+0+....+0+1/2-1/2020

=1010/2020-1/2020=1009/2020

2/3 x - 2/5   = 1/2 x - 1/3

2/3 x - 1/2 x = 2/5 - 1/3        ( chuyển vế đổi dấu )

( 2/3 - 1/2 ) x = 6/15 - 5/15

( 4/6 - 3/6 ) x = 1/15

       1/6       x = 1/15

                    x = 1/15 : 1/6

                    x = 1/15 x 6

                    x = 6/15

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

A, \(2\frac{2}{5}\left(\frac{1}{2}x-0,75\right)=\frac{3}{10}\)

\(=>\frac{2.5+2}{5}\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(=>\frac{1}{2}x-\frac{3}{4}=\frac{3}{10}:\frac{12}{5}=\frac{1}{8}\)

\(=>x=\left(\frac{1}{8}+\frac{3}{4}\right):\frac{1}{2}\)

\(=>x=\frac{7}{4}\)

B, \(\frac{3}{5}-|x-\frac{1}{2}|=25\%\)

\(=>|x-\frac{1}{2}|=\frac{3}{5}-\frac{1}{4}\)

\(=>|x-\frac{1}{2}|=\frac{7}{20}\)

\(=>x-\frac{1}{2}=\frac{7}{20};-\frac{7}{20}\)

TH1: \(x-\frac{1}{2}=\frac{7}{20}=>x=\frac{17}{20}\)

TH2: \(x-\frac{1}{2}=-\frac{7}{20}=>x=\frac{3}{20}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 93 - 3

=> 2x = 90

=> x = 90 : 2

=> x = 45

Vậy x = 45

sai rồi

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\)

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

\(A=\frac{2x}{3\left(2x+3\right)}:2\)

\(\Rightarrow\frac{2x}{3\left(2x+3\right)}:2=\frac{15}{93}\)

\(\frac{2x}{3\left(2x+3\right)}=\frac{15}{93}.2=\frac{30}{93}=\frac{10}{31}\)

\(\frac{2x}{6x+9}=\frac{10}{31}\)

\(\Rightarrow2x.31=10.\left(6x+9\right)\)

\(\Rightarrow62x=60x+90\)

\(62x-60x=90\)

\(2x=90\)

\(x=45\)

Vậy x = 45

Ủng hộ mk nha !!! ^_^

Tui cux k bt nữa tui tính rồi mà tinhs mãi k ra 

Bà bt ln chưa chỉ tôi vs