Câu 2:
b) Phương trình \(x^2-19x+9=0\) có hai nghiệm phân biệt
Theo Vi-et ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-\left(-19\right)}{1}=19\\x_1x_2=\dfrac{9}{1}=9\end{matrix}\right.\)
\(T=\dfrac{x_1\sqrt{x_1}+x_2\sqrt{x_2}}{x^2_1+x^2_2}\)
\(=\dfrac{\left(\sqrt{x_1}\right)^3+\left(\sqrt{x_2}\right)^3}{\left(x_1+x_2\right)^2-2x_1x_2}=\dfrac{\left(\sqrt{x_1}+\sqrt{x_2}\right)\left(x_1-\sqrt{x_1x_2}+x_2\right)}{\left(x_1+x_2\right)^2-2x_1x_2}\)
\(=\dfrac{\sqrt{\left(\sqrt{x_1}+\sqrt{x_2}\right)^2}\cdot\left[\left(x_1+x_2\right)-\sqrt{x_1x_2}\right]}{\left(x_1+x_2\right)^2-2x_1x_2}\)
\(=\dfrac{\sqrt{\left(x_1+x_2\right)+2\sqrt{x_1x_2}}\cdot\left[\left(x_1+x_2\right)-\sqrt{x_1x_2}\right]}{\left(x_1+x_2\right)^2-2x_1x_2}\)
\(=\dfrac{\sqrt{19+2\sqrt{9}}\cdot\left(19-\sqrt{9}\right)}{19^2-2\cdot9}\)
\(=\dfrac{\sqrt{25}\cdot16}{343}=\dfrac{5\cdot16}{343}\)
\(=\dfrac{80}{343}\)
giúp e bài 3b và bài 6 vs ạ
