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\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\))
=2.[(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))
=2.[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)]
2.[(\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x}\))+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.[0+0+...+0+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))
=2.(\(\dfrac{1.x+1-1.2}{2.x+1}\))
=2.(\(\dfrac{x+1-2}{2x}\))=2.\(\dfrac{x-1}{2x}\)=\(\dfrac{2.\left(x-1\right)}{2x}\)=\(\dfrac{2x-2}{2x}\)
\(\dfrac{2x-2}{2x}\)=\(\dfrac{2014}{2016}\)\(\Rightarrow\)(2x-2).2016=2014.2x=4032x-4032=4028x
\(\Rightarrow\)4032x-4028x=4x=4032\(\Rightarrow\)x=4032:4=1008
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)
\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)
D. Tìm x thuộc Z biết
x+(x+1)+(x+2)+....+2016+2017=2017
=> ( x + x + x + ..+ x ) + ( 1 + 2 + 3+...+2016 + 2017 ) = 2017
<=> 2017x + 2035153 = 2017
=> 2017x = -2033136
=> x = -1008
Vậy ...
cảm ơn bạn nhưng bạn có biết những câu hỏi còn lại ko
Ta có:
x biết: 2016 x 2016 - 2015 x 2017 + x = 2016
x = 2015 x 2017 + 2016 - 2016 x 2016
x = 2015 x 2017 + 2016 x (1 - 2016)
x = 2015 x 2017 - 2015 x 2016
x = 2015 x (2017 - 2016)
x = 2015 x 1
x = 2015
Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...
Các bạn làm giúp mình với sao từ nãy giờ chẳng có ai giúp mình vậy?Để mình gợi ý cho các bạn một tí thôi nhé,nhớ giúp mình đấy.
Đáp án là : -1,002482622
Ai có cùng đáp án là làm đúng rồi đấy nếu đúng thì hãy chia sẻ cách làm của các bạn với mình nhé.
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
x=-2016
bn có thể giải thích giúp mk đc k????