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nhân 2 vào 2 vế rồi bạn biến đổi ra( mình lười làm ắ)
tìm được x=50 ắ
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}=2.\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
\(\Rightarrow1-\frac{1}{2x+1}=\frac{98}{99}\)
\(\Rightarrow\frac{2x}{2x+1}=\frac{98}{99}\)
=> 2x = 98
=> x = 98 : 2 = 49
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\) \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)\(\Leftrightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)
\(\Leftrightarrow2x+1=99\)
\(\Rightarrow x=\dfrac{99-1}{2}\)
\(\Rightarrow x=49\)
Vậy \(x=49\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{8}{17}:\dfrac{1}{2}=\dfrac{16}{17}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{16}{17}=\dfrac{1}{17}\)
\(\Rightarrow x+2=17\rightarrow x=15\)
Vậy x = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
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