a, Thay m = 2 ta được \(\left\{{}\begin{matrix}2x+y=1\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

b, \(\Leftrightarrow\left\{{}\begin{matrix}3x=3m-3\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m-1\\y=m-3\end{matrix}\right.\)

Ta có : \(x^2+y^2=m^2-2m+1+m^2-6m+9=2m^2-8m+10\)

\(=2\left(m^2-4m+4-4\right)+10=2\left(m-2\right)^2+2\ge2\forall m\)

Dấu''='' xảy ra khi m =2 

Vậy ...

Ta có: \(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\m\left(2-my\right)-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\2m-m^2y-2y=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\y\left(m^2+2\right)=2m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-\dfrac{m\cdot\left(2m-1\right)}{m^2+2}\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m^2+4-2m^2+m}{m^2+2}=\dfrac{m+4}{m^2+2}\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\)

Tới đây bạn tự làm tiếp nhé

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)

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

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

\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)

\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)

\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)

\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)

Vậy ...

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

nên hệ luôn có nghiệm duy nhất

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

=>\(\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6\left(m+2\right)=6m+12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=3-m+6m+12=5m+15\\x-2y=3-m\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)

Để x>0 và y<0 thì \(\left\{{}\begin{matrix}m+3>0\\m< 0\end{matrix}\right.\)

=>-3<m<0

b: \(A=x^2+y^2=\left(m+3\right)^2+m^2\)

\(=2m^2+6m+9\)

\(=2\left(m^2+3m+\dfrac{9}{2}\right)\)

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

\(=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}>=\dfrac{9}{2}\forall m\)

Dấu '=' xảy ra khi \(m+\dfrac{3}{2}=0\)

=>\(m=-\dfrac{3}{2}\)

Đk để hpt luôn có nghiệm duy nhất (x;y) \(\frac{4}{1}\ne\frac{3}{2}\) (luôn đúng)

\(HPT\Leftrightarrow\hept{\begin{cases}4x-3y=m-10\\4x+8y=12m+12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}11y=11m+22\\x+2y=3m+3\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{11m+22}{11}\\x=3m+3-2y\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{11m+22}{11}\\x=\frac{33m+33-22m-44}{11}\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{11m+22}{11}\\x=\frac{11m-11}{11}\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}x=m-1\\y=m+2\end{cases}}\)

Vậy vơi mọi m thì hpt có nghiệm duy nhất (x;y)=(m-1;m+2)

Ta có:\(x^2+y^2=\left(m-1\right)^2+\left(m+2\right)^2\)

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

\(=2m^2+2m+5=2\left(m^2+m+\frac{5}{2}\right)\)

\(=2\left(m^2+m+\frac{1}{4}+\frac{9}{4}\right)=2\left(m+\frac{1}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\)

Để x²+y² nhỏ nhất <=> \(2\left(m+\frac{1}{2}\right)^2\) nhỏ nhất <=> m+1/2=0 <=> m=-1/2