Đáp án:
Giải thích các bước giải:
$Theo$ $hệ$ $thức$ $Vi - et$$,$ $ta$ $có$ $:$
$\left \{ {{S=x_{1}+x_{2}=-(m-2)}\atop{P= x_1x_{2}=-m+1{}}} \right.$
⇔ $\left \{ {{S=x_{1}+x_{2}=-m+2}\atop{P= x_1x_{2}=-m+1{}}} \right.$
$A$ $=$ $x_{1}$$^{2}$ + $x_{2}$$^{2}$
$⇔$ $A$ $=$ $x_{1}$$^{2}$ $+$ $x_{2}$$^{2}$ $+$ $2$$x_{1}$$x_{2}$ $-$ $2$$x_{1}$$x_{2}$
$⇔$ $A$ $=$ ($x_{1}$$^{2}$ $+$ $2$$x_{1}$$x_{2}$ $+$ $x_{2}$$^{2}$) $-$ $2$$x_{1}$$x_{2}$
$⇔$ $A$ $=$ ($x_{1}$ $+$ $x_{2}$)$^{2}$ $-$ $2$$x_{1}$$x_{2}$
$⇔$ $A$ $=$ ($2$ $-$ $m^{}$)$^{2}$ $-$ $2$.($-$ $m^{}$ $+$ 1)
$⇔$ $A$ $=$ $m^{2}$ $-$ $2$$m + $ $2$
