Giúp e bài 11 từ câu 3-9 với ạa, khó hiểu qa huhuhu

Giải thích các bước giải:

$\begin{aligned}
&\text{Ta có:} \\
&\Delta = (-2)^2 - 4 \cdot 1 \cdot (-1) = 8 > 0 \text{ (Phương trình luôn có hai nghiệm phân biệt)} \\
&x_1 + x_2 = -\dfrac{-2}{1} = 2 \text{} \\
&x_1 x_2 = \dfrac{-1}{1} = -1 \text{} \\
\\
&\text{1) } 5x_1 x_2^2 + 5x_1^2 x_2 \text{} \\
&= 5x_1 x_2(x_2 + x_1) \\
&= 5(-1)(2) \\
&= -10 \\
\\
&\text{2) } x_1^2 + x_2^2 \text{} \\
&= (x_1 + x_2)^2 - 2x_1 x_2 \\
&= 2^2 - 2(-1) \\
&= 4 + 2 \\
&= 6 \\
\\
&\text{3) } 2x_1 x_2^3 + 2x_1^3 x_2 \text{} \\
&= 2x_1 x_2(x_2^2 + x_1^2) \\
&= 2(-1)(6) \\
&= -12 \\
\\
&\text{4) } \dfrac{x_1}{x_2} + \dfrac{x_2}{x_1} \text{} \\
&= \dfrac{x_1^2 + x_2^2}{x_1 x_2} \\
&= \dfrac{6}{-1} \\
&= -6 \\
\\
&\text{5) } \dfrac{3}{x_1} + \dfrac{3}{x_2} \text{} \\
&= \dfrac{3(x_1 + x_2)}{x_1 x_2} \\
&= \dfrac{3(2)}{-1} \\
&= -6 \\
\\
&\text{6) } \dfrac{x_1 + 1}{x_2} + \dfrac{x_2 + 1}{x_1} \text{} \\
&= \dfrac{x_1(x_1 + 1) + x_2(x_2 + 1)}{x_1 x_2} \\
&= \dfrac{x_1^2 + x_1 + x_2^2 + x_2}{x_1 x_2} \\
&= \dfrac{(x_1^2 + x_2^2) + (x_1 + x_2)}{x_1 x_2} \\
&= \dfrac{6 + 2}{-1} \\
&= -8 \\
\\
&\text{7) } \dfrac{x_1}{x_2 + 2} + \dfrac{x_2}{x_1 + 2} \text{} \\
&= \dfrac{x_1(x_1 + 2) + x_2(x_2 + 2)}{(x_2 + 2)(x_1 + 2)} \\
&= \dfrac{x_1^2 + 2x_1 + x_2^2 + 2x_2}{x_1 x_2 + 2x_2 + 2x_1 + 4} \\
&= \dfrac{(x_1^2 + x_2^2) + 2(x_1 + x_2)}{x_1 x_2 + 2(x_1 + x_2) + 4} \\
&= \dfrac{6 + 2(2)}{-1 + 2(2) + 4} \\
&= \dfrac{10}{7} \\
\\
&\text{8) } \dfrac{x_1 - 1}{x_2} + \dfrac{x_2 - 1}{x_1} \text{} \\
&= \dfrac{x_1(x_1 - 1) + x_2(x_2 - 1)}{x_1 x_2} \\
&= \dfrac{x_1^2 - x_1 + x_2^2 - x_2}{x_1 x_2} \\
&= \dfrac{(x_1^2 + x_2^2) - (x_1 + x_2)}{x_1 x_2} \\
&= \dfrac{6 - 2}{-1} \\
&= -4 \\
\\
&\text{9) } \dfrac{x_1^2 + 2}{x_2} + \dfrac{x_2^2 + 2}{x_1} \text{} \\
&= \dfrac{x_1(x_1^2 + 2) + x_2(x_2^2 + 2)}{x_1 x_2} \\
&= \dfrac{x_1^3 + 2x_1 + x_2^3 + 2x_2}{x_1 x_2} \\
&= \dfrac{(x_1^3 + x_2^3) + 2(x_1 + x_2)}{x_1 x_2} \\
&= \dfrac{(x_1 + x_2)^3 - 3x_1 x_2(x_1 + x_2) + 2(x_1 + x_2)}{x_1 x_2} \\
&= \dfrac{2^3 - 3(-1)(2) + 2(2)}{-1} \\
&= \dfrac{8 + 6 + 4}{-1} \\
&= \dfrac{18}{-1} \\
&= -18 \\
\end{aligned}$