Đáp án:
$\frac{-1}{2} <m<2$
Giải thích các bước giải:
$\left \{ {{x - 2y = 1 (1)} \atop {mx + y = 2 (2)}} \right.$
$từ (1) → x = 2y + 1$
$thay$ $vào$ $(2)$ $ta có:$
$m(2y + 1) + y = 2 $
$→ 2my + m + y = 2$
$→ y(2m + 1) = 2 - m $
$→ y = \frac{2-m}{2m+1} (m \neq \frac{-1}{2})$
$→ x = 2.\frac{2-m}{2m+1} + 1 = \frac{4 - 2m}{2m + 1} + 1 = \frac{5}{2m + 1} $
$→ Hệ$ $pt$ $có$ $nghiệm$ $duy$ $nhất$ $(x ; y) = (\frac{5}{2m+1} ; \frac{2-m}{2m+1})$
$Để$ $\left \{ {{x>0} \atop {y >0}} \right.$
$→\left \{ {{\frac{5}{2m+1} > 0} \atop {\frac{2-m}{2m+1} > 0}} \right.$
$→\left \{ {{2m+1>0} \atop {2-m>0}} \right.$
$→\left \{ {{m > \frac{-1}{2}} \atop{m<2}}\right.$
$→ $$\frac{-1}{2} <m<2$
