\(2x^2-3x-1=0\)
Vì \(a\cdot c=2\cdot\left(-1\right)=-2< 0\)
\(\rightarrow\) PT luôn có 2 nghiệm pb trái dấu
Theo Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
Theo đề bài có: \(A=x_1-\dfrac{1}{x_2}+1+x_2-\dfrac{1}{x_1}+1\)
\(A=\left(x_1+x_2\right)-\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)+2\)
\(A=\left(x_1+x_2\right)-\left(\dfrac{x_1+x_2}{x_1x_2}\right)+2\)
\(A=\dfrac{3}{2}-\dfrac{\dfrac{3}{2}}{-\dfrac{1}{2}}+2\)
\(A=\dfrac{3}{2}-\dfrac{3}{2}\cdot\dfrac{2}{-1}+2\)
\(A=\dfrac{3}{2}+3+2=\dfrac{13}{2}\)