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3 tháng 2 2019

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)

\(\Leftrightarrow\) \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)\(\Leftrightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)

\(\Leftrightarrow2x+1=99\)

\(\Rightarrow x=\dfrac{99-1}{2}\)

\(\Rightarrow x=49\)

Vậy \(x=49\)

4 tháng 5 2022

nhân 2 vào 2 vế rồi bạn biến đổi ra( mình lười làm ắ)

tìm được x=50 ắ

4 tháng 5 2022

=49 mà?

23 tháng 11 2021

\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)

23 tháng 11 2021

Em cảm ơn.

1 tháng 6 2016

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 \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)

18 tháng 11 2016

x=49

15 tháng 11 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}=2.\frac{49}{99}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow1-\frac{1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow\frac{2x}{2x+1}=\frac{98}{99}\)

=> 2x = 98

=> x = 98 : 2 = 49

15 tháng 11 2016

cảm ơn bạn rất là nhiều nhé

23 tháng 9 2021

\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)

15 tháng 10 2022

Nguyễn Hoàng Minh cho hỏi 2x + 1 - 1 đâu ra v ạ??

30 tháng 10 2017

mo sach ra ma tim

30 tháng 10 2017

Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1) 
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1) 
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1) 
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1) 
 A = n/(2n + 1)=49/99

Tự tính nha !

17 tháng 8 2018

a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\Leftrightarrow98\left(2x+1\right)=99.2x\)

\(\Leftrightarrow2x=98\Rightarrow x=49\)

b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)

\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)

\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)

\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)

\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)

\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)

\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)

\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)

\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)

\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)

\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)

=>x=2010

30 tháng 10 2017

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2\cdot\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=1-\dfrac{1}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Rightarrow2x+1=99\\ \Leftrightarrow2x=98\\ \Leftrightarrow x=49\)