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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.

28 tháng 3 2017

b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Rightarrow50x\ge0\Rightarrow x\ge0\)

Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)

Thay (1) vào đề bài:

\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)

\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)

\(\Rightarrow49x+\dfrac{16}{99}=50x\)

\(\Rightarrow x=\dfrac{16}{99}\)

Vậy \(x=\dfrac{16}{99}.\)

28 tháng 3 2017

thank bn nhìu nhìu vui

18 tháng 3 2017

Đinh Phương Nguyễn

18 tháng 3 2017

Đinh Phương Nguyễn đây này chú

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

5 tháng 6 2018

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)

\(\Rightarrow\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{8}{17}\)

\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)

\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{8}{17}:\dfrac{1}{2}=\dfrac{16}{17}\)

\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{16}{17}=\dfrac{1}{17}\)

\(\Rightarrow x+2=17\rightarrow x=15\)

Vậy x = 15

6 tháng 6 2018

1/2 nhan voi b/thuc tren bi sai roi

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??

19 tháng 2 2019

Với mọi x ta có :

+) \(\left|x+\dfrac{1}{1.3}\right|\ge0; \)

+) \(\left|x+\dfrac{1}{3.5}\right|\ge0;\)

.....................................

+) \(\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Leftrightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+.......+\left|x+\dfrac{1}{97.99}\right|\ge0\)

\(\Leftrightarrow50x\ge0\)

\(\Leftrightarrow x\ge0\)

Khi \(x\ge0\) ta được :

+) \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3}\)

+) \(\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5}\)

.............................................

+) \(\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\)

\(\Leftrightarrow\left(x+\dfrac{1}{1.3}\right)+\left(x+\dfrac{1}{3.5}\right)+......+\left(x+\dfrac{1}{97.99}\right)=50x\)

\(\Leftrightarrow49x+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{97.99}\right)=50x\)

\(\Leftrightarrow x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{97}-\dfrac{1}{99}\)

\(\Leftrightarrow x=\dfrac{16}{99}\)

Vậy...

23 tháng 10 2017

\(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)

\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)