\(\text{Δ}=\left(-5\right)^2-4\cdot2\cdot1=25-8=17>0\)
Do đó; Phương trình có hai nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{5}{2}\\x_1x_2=\dfrac{1}{2}\end{matrix}\right.\)
\(B=\sqrt{x_1x_2}\left(\sqrt{x_1}+\sqrt{x_2}\right)\)
\(=\sqrt{\dfrac{5}{2}\cdot\dfrac{1}{2}}\cdot\left(x_1+x_2-2\sqrt{x_1x_2}\right)\)
\(=\dfrac{\sqrt{5}}{2}\cdot\left(\dfrac{5}{2}-2\cdot\dfrac{\sqrt{2}}{2}\right)=\dfrac{\sqrt{5}\left(5-2\sqrt{2}\right)}{4}\)
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