\(\dfrac{x^2+1}{x}+\dfrac{x}{x^2+1}=\dfrac{5}{2}\left(1\right)\)(ĐKXĐ: \(x\ne0\))
Đặt \(t=\dfrac{x^2+1}{x}\)
\(\left(1\right):t+\dfrac{1}{t}=\dfrac{5}{2}\\ \Leftrightarrow t+\dfrac{1}{t}-\dfrac{5}{2}=0\\ \Leftrightarrow\dfrac{2t^2+2-5t}{2t}=0\\ \Leftrightarrow2t^2-5t+2=0\\ \Leftrightarrow2t^2-t-4t+2=0\\ \Leftrightarrow t\left(2t-1\right)-2\left(2t-1\right)=0\\ \Leftrightarrow\left(2t-1\right)\left(t-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2t-1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1}{2}\\t=2\end{matrix}\right.\)
Với \(t=\dfrac{1}{2}\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{1}{2}\Leftrightarrow2x^2+2=x\Leftrightarrow2x^2-x+2=0\left(VN\right)\)
Với \(t=2\Leftrightarrow\dfrac{x^2+1}{x}=2\Leftrightarrow x^2+1=2x \Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(TM\right)\)Vậy nghiệm phương trình là \(x=1\)
\(\dfrac{x^2+1}{x}+\dfrac{x}{x^2+1}=\dfrac{5}{2}\left(ĐKXĐ:x\ne0\right)\\ \Rightarrow2\left(x^2+1\right)+2x^2=5x\left(x^2+1\right)\\ \Leftrightarrow2\left(x^4+2x^2+1\right)+2x^2=5x^3+5x\\ \Leftrightarrow2x^4-5x^3+6x^2-5x+2=0\\ \Leftrightarrow2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0\)
\(\Leftrightarrow\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)+\left(3x^2-3x\right)-\left(2x-2\right)=0\\ \Leftrightarrow2x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x^3-3x^2+3x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x^3-2x^2-x^2+x+2x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(2x^3-2x^2)-(x^2-x)+(2x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[2x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x-1\right)\left(2x^2-x+2\right)=0\\ \Leftrightarrow\left(x-1\right)^2\cdot\left(2x^2-x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\2x^2-x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\left(TMĐK\right)\\phương.trình.vô.nghiệm\end{matrix}\right.\)
\(Vậy:S=\left\{1\right\}\)
