Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=1\end{matrix}\right.\)
Ta có: \(C=\dfrac{x_1^4x_2-x_1x_2^4}{\sqrt{x_2}-\sqrt{x_1}}=-\dfrac{x_1x_2\left(x_1^3-x_2^3\right)}{\sqrt{x_1}-\sqrt{x_2}}\)
\(=-\dfrac{x_1x_2\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)}{\sqrt{x_1}-\sqrt{x_2}}\)
Do phương trình có hai nghiệm phân biệt, hay \(x_1\ne x_2\) nên:
\(C=-x_1x_2\left(\sqrt{x_1}+\sqrt{x_2}\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]\)
\(=-x_1x_2\left[\left(x_1+x_2\right)^2-x_1x_2\right]\sqrt{\left(\sqrt{x_1}+\sqrt{x_2}\right)^2}\)
\(=-x_1x_2\left[\left(x_1+x_2\right)^2-x_1x_2\right]\sqrt{\left(x_1+x_2\right)+2\sqrt{x_1x_2}}\)
\(\Rightarrow C=-1\left(4^2-1\right)\sqrt{4+2\sqrt{1}}=-15\sqrt{6}\)
Vậy: \(C=-15\sqrt{6}.\)
