\(5x^2+4x-1=0\)
=>\(x_1\cdot x_2=-\dfrac{1}{5};x_1+x_2=-\dfrac{4}{5}\)
\(A=\left(5x_1-3x_2\right)\left(5x_2-3x_1\right)\)
\(=25x_1x_2-15x_1^2-15x_2^2+9x_1x_2\)
\(=34x_1x_2-15\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\)
\(=34\cdot\dfrac{-1}{5}-15\left[\left(-\dfrac{4}{5}\right)^2-2\cdot\dfrac{-1}{5}\right]\)
\(=-\dfrac{34}{5}-15\cdot\left[\dfrac{16}{25}+\dfrac{2}{5}\right]\)
\(=\dfrac{-34}{5}-15\cdot\dfrac{26}{25}\)
\(=\dfrac{-34}{5}-\dfrac{78}{5}=-\dfrac{112}{5}\)
\(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(-\dfrac{4}{5}\right)^2-2\cdot\dfrac{-1}{5}\)
\(=\dfrac{16}{25}+\dfrac{2}{5}=\dfrac{16+10}{25}=\dfrac{26}{25}\)
\(C=\left|x_1-x_2\right|\)
\(=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{x_1^2+x_2^2-2x_1x_2}=\sqrt{\dfrac{26}{25}-2\cdot\dfrac{-1}{5}}=\dfrac{6}{5}\)
\(D=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-\dfrac{4}{5}\right)^3-3\cdot\dfrac{-1}{5}\cdot\dfrac{-4}{5}\)
\(=-\dfrac{64}{125}-3\cdot\dfrac{4}{25}\)
\(=\dfrac{-64}{125}-\dfrac{12}{25}=\dfrac{-124}{125}\)
