Lời giải:
$\left \{ {{u+v=2} \atop {u.v=-1}} \right.$
<=>$\left \{ {{u=2-v} \atop {(2-v).v=-1}} \right.$
<=>$\left \{ {{u=2-v} \atop {-v^2+2v+1=0(1)}} \right.$(*)
Giải $(1)$,ta được:
$Δ=b^2-4ac=4-4.(-1).1=8>0$.Phương trình có $2$ nghiệm phân biệt.
\(\left[ \begin{array}{l}v_1=\frac{-b+\sqrt{Δ}}{2a}\\v_2=\frac{-b-\sqrt{Δ}}{2a}\end{array} \right.\)
<=>\(\left[ \begin{array}{l}v_1=\frac{-2+\sqrt{8}}{-2}\\v_2=\frac{-2-\sqrt{8}}{-2}\end{array} \right.\)
<=>\(\left[ \begin{array}{l}v_1=1-\sqrt{2}\\v_2=1+\sqrt{2}\end{array} \right.\)
Thay vào (*),ta có:
$\left \{ {{u=2-v} \atop {\left[ \begin{array}{l}v_1=1+\sqrt{2}\\v_2=1-\sqrt{2}\end{array} \right. }} \right.$
<=>$\left \{ {{\left[ \begin{array}{l}u_1=1-\sqrt{2}\\u_2=1+\sqrt{2}\end{array} \right. } \atop {\left[ \begin{array}{l}v_1=1+\sqrt{2}\\v_2=1-\sqrt{2}\end{array} \right. }} \right.$
