Đáp án:
$m=\frac{5}{3}$
Giải thích các bước giải:
${\left\{\begin{aligned}x+my=2\\mx+2y=1\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x=2-my\\m(2-my)+2y=1\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}x=2-my\\2m-m^2y+2y=1\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}x=2-my\\(-m^2+2)y=1-2m\end{aligned}\right.}\\
+)-m^2+2\neq 0\Leftrightarrow m^2\neq 2\Leftrightarrow m\neq \pm \sqrt{2}\\
\Rightarrow {\left\{\begin{aligned}x=2-my\\y=\frac{1-2m}{-m^2+2}\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}x=2-m.\frac{1-2m}{-m^2+2}=\frac{2(-m^2+2)-m(1-2m)}{-m^2+2}\\y=\frac{1-2m}{-m^2+2}\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}x=\frac{-2m^2+4-m+2m^2}{-m^2+2}\\y=\frac{1-2m}{-m^2+2}\end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}x=\frac{4-m}{-m^2+2}\\y=\frac{1-2m}{-m^2+2}\end{aligned}\right.}$
Để phương trình có nghiệm duy nhất thỏa $x+y=0$
$\Rightarrow \frac{4-m}{-m^2+2}+\frac{1-2m}{-m^2+2}=0\\
\Leftrightarrow 4-m+1-2m=0\\
\Leftrightarrow -3m=-5\\
\Leftrightarrow m=\frac{5}{3}$
