Giúp mình câu này với ạ,mình vote 5 sao ạ!!!!

Đáp án:

$\begin{array}{l}
 + x_1^2 + x_2^2 = {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
 + {x_1} - {x_2} =  \pm \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2}} \\
 =  \pm \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
 + x_1^2 - x_2^2\\
 = \left( {{x_1} + {x_2}} \right).\left( {{x_1} - {x_2}} \right)\\
 =  \pm \left( {{x_1} + {x_2}} \right).\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
1)\\
\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ - b}}{a} =  - \frac{1}{2}\\
{x_1}{x_2} = \frac{c}{a} = \frac{{ - 7}}{2}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x_1^2 + x_2^2 = {\left( { - \frac{1}{2}} \right)^2} - 2.\left( { - \frac{7}{2}} \right) = \frac{{29}}{4}\\
{x_1} - {x_2} =  \pm \sqrt {{{\left( { - \frac{1}{2}} \right)}^2} - 4.\left( { - \frac{7}{2}} \right)}  =  \pm \frac{{\sqrt {57} }}{2}\\
x_1^2 - x_2^2 =  - \frac{1}{2}. \pm \frac{{\sqrt {57} }}{2} =  \pm \frac{{\sqrt {57} }}{4}
\end{array} \right.\\
2)\\
\left\{ \begin{array}{l}
{x_1} + {x_2} = 1\\
{x_1}{x_2} =  - 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x_1^2 + x_2^2 = 1 - 2.\left( { - 1} \right) = 3\\
{x_1} - {x_2} =  \pm \sqrt {{1^2} - 4.\left( { - 1} \right)}  =  \pm \sqrt 5 \\
x_1^2 - x_2^2 =  \pm \sqrt 5 
\end{array} \right.
\end{array}$