Cho hệ phương trình: $\begin{cases}mx+4y=m^2+4 \\ x+(m+3)y=2m+3 \end{cases}$
Với giá trị nào của $m$ thì hệ có nghiệm duy nhất thỏa mãn điều kiện $x\geqslant y$.
$D=|\begin{matrix} m & 4\\ 1 & m+3 \end{matrix}|=m^2+3m-4=0\Leftrightarrow m=1, m=-4$
.
$D_x=|\begin{matrix} m^2+4 & 4\\ 2m+3 & m+3 \end{matrix}|=m(m^2+3m-4)$

$D_y=|\begin{matrix} m & m^2+4\\ 1 & 2m+3 \end{matrix}|=m^2+3m-4$

Hệ có nghiệm duy nhất nếu $D\neq0\Leftrightarrow m\neq1, m\neq-4$. Khi đó:
$$x=\frac{D_x}{D}=m,             y=\frac{D_y}{D}=1$$
Điều kiện $x\geq y\Leftrightarrow m\geq 1$.

Do đó, điều kiện $m\neq1,m\neq-4$ nên ta kết luận: Với $m>1$ hệ có nghiệm duy nhất thỏa mãn $x\geq y$
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