a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
c) Ta có: \(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\)
\(\Leftrightarrow\dfrac{2-x}{2017}+1=\dfrac{1-x}{2018}+1-\dfrac{x}{2019}+1\)
\(\Leftrightarrow\dfrac{2-x}{2017}+1=\left(\dfrac{1-x}{2018}+1\right)-\left(\dfrac{x}{2019}-1\right)\)
\(\Leftrightarrow\dfrac{2-x+2017}{2017}=\dfrac{1-x+2018}{2018}-\dfrac{x-2019}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}-\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)
\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
\(\Leftrightarrow2019-x=0\)(vì \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\ne0\))
\(\Leftrightarrow x=2019\)
Vậy nghiệm của phương trình là \(x=2019\)
