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\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2017-1=2016\)
Vậy x = 2016
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2016}{2017}\)
1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)
\(\dfrac{3}{4}\)+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)
\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{2013}{8068}\)
Bn tự lm tiếp nhé!!! Sorry mk đang vội
1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016
<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016
<=> 1 - 1/x+1 = 2015/2016
<=> 1/x+1 = 1/2016
<=> x + 1 = 2016
<=> x = 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)
=>1-1/2+1/2-1/3+...+1/x-1/(x+1)=2022/2021
=>1-1/(x+1)=2022/2021
=>1/(x+1)=-1/2021=1/-2021
=>x+1=-2021
=>x=-2022
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
(x-1) là ước của 6
\(\Rightarrow\left(x-1\right)\in\left\{1;2;3;6;-1;-2;-3;-6\right\}.\)
\(\Rightarrow x\in\left\{2;3;4;7;0;-1;-2;-5\right\}\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)
\(\left(2x-1\right)=-\frac{4}{63}\)
2x= -4/63 + 1
2x = 59/63
x = 59/63 : 2
x = 59/126
1/3:(2.x-1)=-5-1/4
1/3:(2.x-1)=-21/4
2.x-1=1/3:-21/4
2.x-1=-4/63
2.x=-4/63+1
2.x=\(3\frac{59}{63}\)
x=\(3\frac{59}{63}\):2
x=\(1\frac{61}{63}\)
đặt A=.....
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)=\(\frac{2016}{2017}\)
=\(1-\frac{1}{x+1}=\frac{2016}{2017}\)
=\(\frac{x}{x+1}=\frac{2016}{2017}\)
=>x=2016
vậy..............