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Nhân cả 2 vế với 3 ta có:
\(pt\Leftrightarrow2x-\dfrac{6}{y}=1\Leftrightarrow2x=1+\dfrac{6}{y}\)
Nhận thấy rằng 2x là số nguyên, 1 là số nguyên nên \(\dfrac{6}{y}\) cũng là số nguyên
=> y ∈ Ư(6) = {\(\pm\)1; \(\pm\)2; \(\pm\)3; \(\pm\)6}
Mà 2x là số chẵn => \(1+\dfrac{6}{y}\) là số chẵn => y ∈ {\(\pm\)2; \(\pm\)6}
+) \(y=-6\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{-6}\right)=0\)
+) \(y=-2\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{-2}\right)=-1\)
+) \(y=2\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{2}\right)=2\)
+) \(y=6\Rightarrow x=\dfrac{1}{2}\left(1+\dfrac{6}{6}\right)=1\)
Ta có : \(A=\dfrac{x^2+2x+1-4x-4+4}{x+1}\)
\(=\dfrac{\left(x+1\right)^2-4\left(x+1\right)+4}{x+1}=x+1-4+\dfrac{4}{x+1}\)
- Để A là số nguyên
\(\Leftrightarrow x+1\inƯ_{\left(4\right)}\) ( Do x là số nguyên )
\(\Leftrightarrow x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;1;-3;3;-5\right\}\)
Vậy ....
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+......+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{3}{16}\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{3}{16}\)
\(\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{3}{16}:\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{3}{8}\)
\(\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{3}{8}\)
\(\dfrac{1}{2x}=\dfrac{1}{8}\)
⇒x=8:2=4
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(2017>2016>2015>2014\) nên
\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\Rightarrow x=1009\)
Vậy...........
Chúc bạn học tốt!!!
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)
\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\)
\(\Rightarrow2x=2018-0\)
\(\Rightarrow2x=2018\)
\(\Rightarrow x=2018:2\)
\(\Rightarrow x=1009\)
dễ mà