\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2a-1\\\left(x+y\right)^2-2xy=a^2+2a-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2a-1\\2xy=\left(2a-1\right)^2-\left(a^2+2a-3\right)\end{matrix}\right.\)

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

Hệ pt đã cho có nghiệm \(\Leftrightarrow\left(2a-1\right)^2\ge4\left(\frac{3a^2-6a+4}{2}\right)\)

\(\Leftrightarrow4a^2-4a+1\ge6a^2-12a+8\)

\(\Leftrightarrow2a^2-8a+7\le0\Rightarrow\frac{4-\sqrt{2}}{2}\le a\le\frac{4+\sqrt{2}}{2}\)

Khi đó: \(f\left(a\right)=xy=\frac{3a^2-6a+4}{2}=\frac{3}{2}a^2-3a+2\)

Xét \(f\left(a\right)\) trên \(\left[\frac{4-\sqrt{2}}{2};\frac{4+\sqrt{2}}{2}\right]\)

\(\frac{3}{2}>0;\) \(\frac{3}{2.\frac{3}{2}}=1< \frac{4-\sqrt{2}}{2}\Rightarrow f\left(a\right)\) đồng biến trên \(\left[\frac{4-\sqrt{2}}{2};\frac{4+\sqrt{2}}{2}\right]\)

\(\Rightarrow f\left(a\right)_{min}=f\left(\frac{4-\sqrt{2}}{2}\right)=\frac{11-6\sqrt{2}}{4}\)

Từ phương trình trên , suy ra :

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

\(\Leftrightarrow4a^2-4a+1=\left(a^2-2a-3\right)+2xy\)

\(\Leftrightarrow3a^2-2a+4=2xy\)

\(\Leftrightarrow3\left(a^2-\frac{2}{3}a+\frac{4}{3}\right)=2xy\)

\(\Leftrightarrow3\left(a^2-\frac{2}{3}a+\frac{1}{9}\right)+\frac{11}{3}=2xy\)

\(\Leftrightarrow3\left(a-\frac{1}{3}\right)^2+\frac{11}{3}=2xy\)

Nhận thấy \(VT\ge\frac{11}{3}\)suy ra  \(2xy\ge\frac{11}{3}\) => \(xy\ge\frac{11}{6}\)

Vậy Min(xy) = 11/6 <=> a = 1/3

giúp mik đc ko, mikk cần gấp

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

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

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

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

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

Để x+y>0 thì \(\dfrac{m+2}{2m-1}+\dfrac{m^2-3m}{2m-1}>0\)

\(\Leftrightarrow\dfrac{m+2+m^2-3m}{2m-1}>0\)

\(\Leftrightarrow\dfrac{m^2-2m+2}{2m-1}>0\)

mà \(m^2-2m+2>0\forall m\)

nên 2m-1>0

\(\Leftrightarrow2m>1\)

hay \(m>\dfrac{1}{2}\)

Vậy: Để hệ phương trình có nghiệm duy nhất thỏa mãn x+y>0 thì \(m>\dfrac{1}{2}\)

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

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

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

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

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

Hpt có nghiệm duy nhất \(\Leftrightarrow\) 2m - 1 \(\ne\) 0 \(\Leftrightarrow\) m \(\ne\) \(\dfrac{1}{2}\)

Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{2+m}{2m-1}\\y=m-m.\dfrac{2+m}{2m-1}\end{matrix}\right.\)

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

Vậy hpt có nghiệm duy nhất (x; y) = ...

Ta có: x + y > 0

\(\Leftrightarrow\) \(\dfrac{m^2-2m+2}{2m-1}>0\)

\(\Leftrightarrow\) \(\dfrac{\left(m-1\right)^2+1}{2m-1}\) > 0

\(\Leftrightarrow\) 2m - 1 > 0 (vì (m - 1)² + 1 > 0 với mọi m)

\(\Leftrightarrow\) 2m > 1

\(\Leftrightarrow\) m > \(\dfrac{1}{2}\)

Kết hợp với m \(\ne\) \(\dfrac{1}{2}\) ta có: m > \(\dfrac{1}{2}\) thì hpt có nghiệm duy nhất (x;y) thỏa mãn x + y > 0

Vậy m > \(\dfrac{1}{2}\)

Chúc bn học tốt! (Chắc đúng :D)