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5 tháng 5 2017

\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)

\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)

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

\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)

\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)

\(x=\dfrac{198}{99}\)

Vậy \(x=\dfrac{198}{99}\)

11 tháng 3 2023

\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)

11 tháng 3 2023

\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)

\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)

và cứ như thế đến số cuối

 

21 tháng 3 2017

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

21 tháng 3 2017

với -1 nữa nha

Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)

\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)

\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)

\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)

Suy ra: x+2=82

hay x=80

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.

19 tháng 3 2023

\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)

\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(B=\dfrac{1}{1}-\dfrac{1}{99}\)

\(B=\dfrac{99}{99}-\dfrac{1}{99}\)

\(B=\dfrac{98}{99}\)

#YVA

22 tháng 3 2023

B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)

B=\(\dfrac{98}{99}:2\)

B=\(\dfrac{49}{99}\)

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

28 tháng 3 2017

\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{97.99}\)


Ta thấy:\(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{1.3 }\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
............\(\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{2}{97.99}\)
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{99}\) =\(\dfrac{1}{1}-\dfrac{1}{99}\)

=\(\dfrac{99}{99}-\dfrac{1}{99}\)

=\(\dfrac{98}{99}\)
Vậy A=\(\dfrac{98}{99}\)

28 tháng 3 2017

A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

A = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

A = \(1-\dfrac{1}{99}\)

A = \(\dfrac{98}{99}\)

6 tháng 4 2018

Ta có :

\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)

\(\dfrac{1}{2}\)(\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)

\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\)=\(\dfrac{40}{41}\)

\(\dfrac{1}{x+2}\)=\(\dfrac{1}{3}\)-\(\dfrac{40}{41}\)

6 tháng 4 2018

1/3-1/1+1/7-1/5+1/9-1/7...

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