Cho hệ phương trình

- Hướng dẫn giải

Ta có:

$D=\left|\begin{array}{cc}m& -1\\ 1& -m\end{array}\right|=-m^2+1=\left(1-m\right)\left(1+m\right)$

$D_x=\left|\begin{array}{cc}2m& -1\\ m+1& -m\end{array}\right|=-2m^2+m+1=\left(2m+1\right)\left(1-m\right)$

$D_y=\left|\begin{array}{cc}m& 2m\\ 1& m+1\end{array}\right|=m^2-m=m\left(m-1\right)$

Hệ phương trình có nghiệm duy nhất $\iff D\ne 0$

$\iff -m^2+1\ne 0\iff m\ne \pm 1\Rightarrow \left\{\begin{array}{c}x=\frac{D_x}{D}=\frac{2m+1}{m+1}\\ y=\frac{D_y}{D}=\frac{-m}{m+1}\end{array}\right.$

Khi đó: $p=x.y=\frac{-m\left(2m+1\right)}{{\left(m+1\right)}^2}=\frac{-2\left(m^2+2m+1\right)}{{\left(m+1\right)}^2}+\frac{3\left(m+1\right)}{{\left(m+1\right)}^2}-\frac{1}{{\left(m+1\right)}^2}$

$=-2+\frac{3}{m+1}-\frac{1}{{\left(m+1\right)}^2}$

Đặt $\frac{1}{m+1}=t$

$\Rightarrow \frac{1}{{(m+1)}^2}=t^2\Rightarrow P=-2+3t-t^2=-{\left(t-\frac{3}{2}\right)}^2+\frac{1}{4}\le \frac{1}{4}$$\Rightarrow P_{m\text{ax}}=\frac{1}{4}$

Dấu “=” xảy ra $\iff t=\frac{3}{2}\iff \frac{1}{m+1}=\frac{3}{2}\iff m=-\frac{1}{3}$

Đáp án cần chọn là: D