\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{45.46}=-2\)

\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{45.46}\right)=-2\)

\(\frac{1}{2.x}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{45}-\frac{1}{46}\right)=-2\)

\(\frac{1}{2.x}-\left(1-\frac{1}{46}\right)\)

\(\frac{1}{2.x}-\frac{45}{46}=-2\)

\(\frac{1}{2.x}=-2+\frac{45}{46}\)

\(\frac{1}{2.x}=\frac{-47}{46}\)

\(2x=\frac{46}{-47}\)

\(x=\frac{46}{-47}:2=\frac{-23}{47}\)

\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-......-\frac{1}{45.46}=-2\)2

\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{45.46}\right)=-2\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{45.46}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{45}-\frac{1}{46}\)

\(A=1-\frac{1}{46}=\frac{45}{46}\)

Ta có: \(\frac{1}{2.x}-\frac{45}{46}=-2\)

\(\frac{1}{2.x}=\frac{-47}{46}\)

\(\frac{-47}{-94.x}=\frac{-47}{46}\)

\(\Rightarrow x=\frac{-23}{47}\)

\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\frac{-1^2}{2}\)

\(=\frac{75}{100}-\frac{3}{2}+\frac{5}{10}.\frac{12}{5}-\frac{1}{2}\)

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

\(=\frac{15}{20}-\frac{30}{20}+\frac{24}{20}-\frac{10}{20}\)

\(=\frac{15-30+24-10}{20}\)

\(=\frac{-1}{20}\).

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)

Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3

=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3

=>1-1/x+1=2/3

=>1/x+1=1/3

=>3=x+1

=>x=2

Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

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

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

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

=>\(\frac{1}{x+1}=\frac{1}{3}\)

=>\(x+1=3\)

=>\(x=2\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{2}{3}\)

\(\Rightarrow1-\frac{1}{x-1}=\frac{2}{3}\)

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

\(\Rightarrow x=3+1=4\)

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

 Tham khảo:

A=1.2+2.3+3.4+...+2013.2014

3A = 1.2.3 + 2.3.3 + 3.4.3 +...+ 2013.2014.3

Mà: 1.2.3 = 1.2.3

2.3.3 = 2.3.4 - 2.3.1

3.4.3 = 3.4.5 - 3.4.2

2012.2013.3  = 2012.2013.2014 - 2012.2013.2011

2013.2014.3 = 2013.2014.2015 - 2013.2014.2012

=> 3S = 2013.2014.2015

=> A = 2013.2014.2015 / 3 = 2723058910