Bài 1:

Ta có:

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

b, Đặt  \(A=\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

\(\Rightarrow\frac{2}{5}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

Từ (a) \(\Rightarrow\frac{2}{5}A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}:\frac{2}{5}=\frac{100}{101}.\text{5/2}=\frac{250}{101}\)

Bài 2:

Đặt \(\left(2n+1;3n+2\right)=d\left(d\inℕ^∗\right)\)

\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}}\)

\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)\Rightarrow d=1\)

\(\Rightarrow\left(2n+1;3n+2\right)=1\)

\(\Rightarrow\frac{2n+1}{3n+2}\)là phân số tối giản

1.          Giải 

a,  \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=2.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)

\(=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

b,   \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(=5.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{5.100}{2.101}=\frac{500}{202}=\frac{250}{101}\)

2.    Giải 

Gọi ước chung lớn nhất của 2n + 1 và 3n + 2 là d (d thuộc N*) 

=> 2n + 1 \(⋮\)d ; 3n + 2 \(⋮\)d 

=> 3(2n + 1) \(⋮\)d ; 2(3n + 2) \(⋮\)d

=> 6n + 3 \(⋮\)d , 6n + 4 \(⋮\)d 

=> (6n + 4) - (6n + 3) \(⋮\)d 

=> 1 \(⋮\)d 

=> d = 1 

Vậy \(\frac{2n+1}{3n+2}\)là phân số tối giản 

a) pt => 2x-x=-25+5(chuyển vế đổi dấu) =>x=-20

b)pt=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\)=\(\frac{2016}{2017}\)

      =>\(1-\frac{1}{2x+1}=\frac{2016}{2017}\)=>\(\frac{2x}{2x+1}=\frac{2016}{2017}\). Nhân chéo => x=1008

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

\(\Rightarrow A< \frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{X\left(X+2\right)}\)

\(\frac{1}{2}.\left(\frac{1}{1.3}+...+\frac{1}{X\left(X+2\right)}\right)\)= \(\frac{16}{34}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{X}-\frac{1}{X+2}\right)\)

=15

TA CÓ :    1/1.3 + 1/3.5 + 1/5.7 +... + 1/X(X+2) = 8/17

        =>    2/1.3 + 2/3.5 + 2/5.7 +... + 2/X(X+2) = 8/17 . 2 = 16/17

      <=>                       1 - 1/X+2                      = 16/17

                       X+2/X+2 - 1/X+2                       = 16/17

                      X+2 -1/X+2                                = 16/17

           => X+2 -1 =16 VÀ X+2 = 17

           => X = 15

<=>2-2/3+2/3-2/5........+2n-2n+2<2015/2016

<=>2-2n+2<2015/2016

=>n+2=1/2016

=>n=2014

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{n\left(n+2\right)}\)<\(\frac{2015}{2016}\)

VT=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{5}-\frac{1}{n+2}\)=\(1-\frac{1}{n+2}\)

Ta có:\(1-\frac{1}{n+2}=\frac{2015}{2016}\Rightarrow\)\(\frac{1}{n+2}=1-\frac{2015}{2016}\)

\(\Rightarrow\)\(\frac{1}{n+2}=\frac{1}{2016}=n+2=2016\)

\(\Rightarrow\)\(n=2014\)

Vậy\(n=2014\)

Gọi \(A=\frac{1005}{2011}\)

A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)

A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2

A . 2=1/1+(1/3-1/3)+(1/5-1/5)+..............+(1/x-1/x)-1/x+2

A . 2=1/1-1/x+2

Suy gia:1005/2011 . 2=1/1-1/x+2

             2010/2011    =1/1-1/x+2

             1/x+2           =1/1-2010/2011

              1/x+2          =1/2011

Suy gia:x+2=2011

            x    =2011-2

            x    =2009

\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(100=2x+4\)

\(\Leftrightarrow\)\(2x=96\)

\(\Leftrightarrow\)\(48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)

\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(49=x+1\)

\(\Leftrightarrow\)\(x=48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

mình cũng đang bí