\(\text{Δ}=\left(-5\right)^2-4\cdot2\cdot1=25-8=17>0\)
=>Phương trình có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{2}\\x_1x_2=\dfrac{c}{a}=\dfrac{1}{2}\end{matrix}\right.\)
\(\left(\sqrt{x_2}+\sqrt{x_1}\right)^2=x_1+x_2+2\sqrt{x_1x_2}\)
\(=\dfrac{5}{2}+2\sqrt{\dfrac{1}{2}}=\dfrac{5}{2}+\sqrt{2}=\dfrac{5+2\sqrt{2}}{2}\)
=>\(\sqrt{x_2}+\sqrt{x_1}=\sqrt{\dfrac{5+2\sqrt{2}}{2}}=\dfrac{\sqrt{10+4\sqrt{2}}}{2}\)
\(B=x_1\sqrt{x_2}+x_2\sqrt{x_1}\)
\(=\sqrt{x_1x_2}\left(\sqrt{x_1}+\sqrt{x_2}\right)\)
\(=\sqrt{\dfrac{1}{2}}\cdot\sqrt{\dfrac{5+2\sqrt{2}}{2}}=\dfrac{\sqrt{5+2\sqrt{2}}}{2}\)