Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne-\dfrac{1}{m}\)
=>\(-m^2\ne2\)(luôn đúng)
\(\left\{{}\begin{matrix}mx-y=m^2\\2x+my=m^2+2m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-m^2\\2x+m\left(mx-m^2\right)=m^2+2m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-m^2\\2x+m^2x=m^3+m^2+2m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-m^2\\x\left(m^2+2\right)=\left(m+1\right)\left(m^2+2\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1\\y=mx-m^2=m\left(m+1\right)-m^2=m\end{matrix}\right.\)
Đặt \(A=x^2+3y+4\)
\(=\left(m+1\right)^2+3m+4\)
\(=m^2+5m+5\)
\(=m^2+2\cdot m\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}\)
\(=\left(m+\dfrac{5}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi m=-5/2
