\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1993}\)

\(\Leftrightarrow x=1992\)

\(\text{Vậy x = 1992 }\)

Câu 1:

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}.\)

\(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)

\(\frac{1}{2.3}.2+\frac{1}{3.4}.2+\frac{1}{4.5}.2+...+\frac{1}{x.\left(x+1\right)}.2=\frac{1991}{1993}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1991}{1993}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)

...

e tự tính nốt nha

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(\Leftrightarrow x=1993-1\)

\(\Leftrightarrow x=1992\)

Vậy x = 1992

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

a)Ta có   \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)

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

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

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

\(\Leftrightarrow3\left(x+\frac{1}{2}\right)=1\)

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

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

\(\Leftrightarrow x=-\frac{1}{6}\)

x=2009 dễ mà

mk làm câu c cho nó dễ

c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010

=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010

=1-1/x+1=2009/2010

=1/x+1=1-2009/2010

=1/x+1=1/2010

=) x+1=2010

x         =2010-1

x         =2009

Đề cho dài :v. Lần sau đăng từ từ nhé bạn, hôm qua đến giờ mình giải không hết đó =(((

a) \(\frac{1}{2}.x-\frac{3}{4}.x-\frac{7}{3}=-\frac{5}{6}=\frac{-5}{6}\)

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

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{3}{2}\Leftrightarrow x.\frac{-1}{4}=\frac{3}{2}\)

\(x=\frac{3}{2}:\frac{-1}{4}=-6\)

b) \(\frac{4}{5}.x-x-\frac{3}{2}.x+\frac{4}{3}=\frac{1}{2}-\frac{6}{5}=-\frac{7}{10}\)

\(\Leftrightarrow x\left(\frac{4}{5}-\frac{3}{2}.\frac{4}{3}\right)=x\left(\frac{4}{5}-2\right)=-\frac{7}{10}\)

\(\Leftrightarrow x.\frac{-6}{5}=-\frac{7}{10}\)

\(x=-\frac{7}{10}:\frac{-6}{5}=\frac{7}{12}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(=1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\frac{1}{x+1}=1-\frac{2009}{2010}=\frac{1}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010-1}=\frac{1}{2009}\). Vậy x= 2009

d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}=\frac{4023}{2015}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{4023}{2015}:2=\frac{4023}{4030}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{4023}{4030}=\frac{-1004}{2015}=\frac{1004}{-2015}\)

\(x+1=\hept{\begin{cases}2015\\-2015\end{cases}}\Rightarrow x=\hept{\begin{cases}2014\\-2016\end{cases}}\)

e) Bạn tự làm, nhiều quá mình làm không hết

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).