ĐK : x ≠ 0
\(\dfrac{x+1}{x}=3\)
=> \(3x=x+1\)
=> \(2x=1\)
=> \(x=\dfrac{1}{2}\)
=> \(x^2=\dfrac{1}{4}\)
=> \(\dfrac{1}{x^2}=4\)
Khi đó
\(\dfrac{x^2+1}{x^2}=1+\dfrac{1}{x^2}=1+4=5\)
\(x+\dfrac{1}{x}=3\Rightarrow x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=9\Rightarrow x^2+\dfrac{1}{x^2}=7\)