Phải là dấu ngoặc nhọn chứ=0

\(\left\{{}\begin{matrix}2x+3y-6< 0\\x\ge0\\2x-3y-1\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3y-6\le2x+3y-6< 0\\x\ge0\\-3y-1\le2x-3y-1\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3y-6< 0\\-3y-1\le0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y< 2\\y\ge-\dfrac{1}{3}\\x\ge0\end{matrix}\right.\)

=> Miền nghiệm là \([0;2)\)

Vì \(\dfrac{1}{2}\ne\dfrac{-2}{3}\)

nên hệ luôn có nghiệm duy nhất

a: \(\left\{{}\begin{matrix}x-2y=-3m-4\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-6m-8\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x-3y=-6m-8-8m+1\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=-14m-7\\2x=8m-1-3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\2x=8m-1-6m-3=2m-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\x=m-2\end{matrix}\right.\)

Đặt \(A=y^2+3x-1\)

\(=\left(2m+1\right)^2+3\left(m-2\right)-1\)

\(=4m^2+4m+1+3m-6-1\)

\(=4m^2+7m-6\)

\(=4\left(m^2+\dfrac{7}{4}m-\dfrac{3}{2}\right)\)

\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{8}+\dfrac{49}{64}-\dfrac{145}{64}\right)\)

\(=4\left(m+\dfrac{7}{8}\right)^2-\dfrac{145}{16}>=-\dfrac{145}{16}\)
Dấu '=' xảy ra khi m=-7/8

b: Đặt B=x^2-y^2

\(=\left(m-2\right)^2-\left(2m+1\right)^2\)

\(=m^2-4m+4-4m^2-4m-1\)

\(=-3m^2-8m+3\)

\(=-3\left(m^2+\dfrac{8}{3}m-1\right)\)

\(=-3\left(m^2+2\cdot m\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{25}{9}\right)\)

\(=-3\left(m+\dfrac{4}{3}\right)^2+\dfrac{25}{3}< =\dfrac{25}{3}\)

Dấu '=' xảy ra khi m=-4/3

a: Vì \(\dfrac{1}{2}\ne\dfrac{-2}{1}=-2\)

nên hệ phương trình luôn có nghiệm duy nhất

b: \(\left\{{}\begin{matrix}x-2y=4-m\\2x+y=8m+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2y=4-m\\4x+2y=16m+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+4x=4-m+16m+6=15m+10\\2x+y=8m+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=15m+10\\y=8m+3-2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3m+2\\y=8m+3-6m-4=2m-1\end{matrix}\right.\)

Đặt \(A=x^2+y^2\)

\(=\left(3m+2\right)^2+\left(2m-1\right)^2\)

\(=9m^2+12m+4+4m^2-4m+1\)

\(=13m^2+8m+5\)

\(=13\left(m^2+\dfrac{8}{13}m+\dfrac{5}{13}\right)\)

\(=13\left(m^2+2\cdot m\cdot\dfrac{4}{13}+\dfrac{16}{169}+\dfrac{49}{169}\right)\)

\(=13\left(m+\dfrac{4}{13}\right)^2+\dfrac{49}{13}>=\dfrac{49}{13}\forall m\)

Dấu '=' xảy ra khi \(m+\dfrac{4}{13}=0\)

=>\(m=-\dfrac{4}{13}\)