1) Thay m = 4 (Tm) vào HPT, ta có:
\(\left\{{}\begin{matrix}2x-3y=10\left(1\right)\\x-y=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-3y=10\\2x-2y=8\end{matrix}\right.\)
\(\Leftrightarrow\) (2x-3y) - (2x-2y) = 10 - 8
\(\Leftrightarrow\) -y = 2 \(\Leftrightarrow\) y = -2
Thay y = -2 vào phương trình (1), ta có:
2x - 3.(-2) = 10
\(\Leftrightarrow x=2\)
KL: Vậy HPT có nghiệm (x;y) = (2;-2)
2)
\(\left\{{}\begin{matrix}2x-3y=2\sqrt{m}+6\\x-y=\sqrt{m}+2\left(2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=2\sqrt{m}+6\\2x-2y=2\sqrt{m}+4\end{matrix}\right.\)
\(\Leftrightarrow\left(2x-3y\right)-\left(2x-2y\right)=2\sqrt{m}+6-2\sqrt{m}-4\)
\(\Leftrightarrow-y=2\Leftrightarrow y=-2\)
Thay y = -2 vào phương trình (2), ta có
\(x+2=\sqrt{m}+2\Leftrightarrow x=\sqrt{m}\)
Để \(x+y< -1\)
\(\Leftrightarrow\sqrt{m}-2< -1\)
\(\Leftrightarrow\sqrt{m}< 1\)
\(\Leftrightarrow0\le m< 1\)
