a: \(4x^2-3x-1=0\)
=>\(4x^2-4x+x-1=0\)
=>(x-1)(4x+1)=0
=>\(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{2}\\x_1x_2=\dfrac{c}{a}=\dfrac{1}{2}\end{matrix}\right.\)
\(A=\dfrac{2x_1^2+5x_2}{\left|x_1-x_2\right|}\)
\(=\dfrac{2x_1^2+2x_2\cdot\left(x_1+x_2\right)}{\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}}=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{\sqrt{\dfrac{25}{4}-4\cdot\dfrac{1}{2}}}\)
\(=\dfrac{2\left(x_1+x_2\right)^2-4x_1x_2+2x_1x_2}{\sqrt{\dfrac{25}{4}-2}}=\dfrac{2\left(\dfrac{5}{2}\right)^2-2\cdot\dfrac{1}{2}}{\sqrt{\dfrac{17}{4}}}\)
\(=\dfrac{2\cdot\dfrac{25}{4}-1}{\dfrac{\sqrt{17}}{2}}=\dfrac{23}{2}:\dfrac{\sqrt{17}}{2}=\dfrac{23}{\sqrt{17}}\)
