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

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

\(A=\left(m+3\right)^2+m^2=2m^2+6m+9=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)

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

Bài 1.

\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

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

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

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

Vì 1/2<>1/3

nên hệ luôn có nghiệm duy nhất

x+y=2 và 2x+3y=m

=>2x+2y=4 và 2x+3y=m

=>-y=4-m và x+y=2

=>y=m-4 và x=2-y=2-m+4=6-m

x+2y<5

=>6-m+2m-8<5

=>m-2<5

=>m<7

=>Có 6 số nguyên dương thỏa mãn

Lấy pt 1 cộng vế với vế của pt 2 ta được

\(2x+y+x-y=m+2+m\Leftrightarrow3x=2m+2\Leftrightarrow x=\dfrac{2m+2}{3}\)

từ pt 2 ta suy ra \(y=\dfrac{-m+2}{3}\)

Để hpt có nghiệm \(x_0,y_0\) thoả mãn đk đề bài thì \(\dfrac{-m+2}{3}+\dfrac{2m+2}{3}=3\Leftrightarrow\dfrac{m+4}{3}=3\Leftrightarrow m=5\)

Vậy ..........

m=5

lay tren tru duoi dc x+2y=2 ma x+y=3 suy ra y=-1 (moi tham gia chua ranh )

=( U GAY

\(\left\{{}\begin{matrix}x_0-my_0=2-4m\\mx_0+y_0=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0-2=m\left(y_0-4\right)\\y_0-1=m\left(3-x_0\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x_0-2\right)\left(3-x_0\right)=m\left(y_0-4\right)\left(3-x_0\right)\\\left(y_0-1\right)\left(y_0-4\right)=m\left(y_0-4\right)\left(3-x_0\right)\end{matrix}\right.\)

\(\Rightarrow\left(x_0-2\right)\left(3-x_0\right)=\left(y_0-1\right)\left(y_0-4\right)\)

\(\left\{{}\begin{matrix}mx+y=5\\2x-y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)x=3\\2x-y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{m+2}\\\frac{6}{m+2}-y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{m+2}\\y=\frac{10+2m}{m+2}\end{matrix}\right.\)

\(\Rightarrow x+y=\frac{3}{m+2}+\frac{10+2m}{m+2}=\frac{13+2m}{m+2}\)

\(\Leftrightarrow\frac{13+2m}{m+2}=1\Leftrightarrow13+2m=m+2\)

\(\Leftrightarrow m=-11\)