$\begin{cases}x+my=m+1\\y+mx=3m-1\\\end{cases}$
$\Leftrightarrow\begin{cases}x=m+1-my\\y+m(m+1-my)=3m-1\\\end{cases}$
$\Leftrightarrow\begin{cases}x=m+1-my\\y-my^2+m^2+m=3m-1\\\end{cases}$
$\Leftrightarrow\begin{cases}x=m+1-my\\y(m^2-1)=m^2-2m+1\\\end{cases}$
Để HPT có nghiệm duy nhất thì $m^2-1 \neq 0\\\Leftrightarrow m \ne \pm1$
$\Leftrightarrow\begin{cases}y=\dfrac{(m-1)^2}{(m-1)(m+1)}=\dfrac{m-1}{m+1}\\x=m+1-my=\dfrac{(m+1)^2-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\\\end{cases}$
$\Rightarrow xy=\dfrac{(3m+1)(m-1)}{(m+1)^2}$
$=\dfrac{3m^2-2m-1}{(m+1)^2}$
Xét $xy+1$
$=\dfrac{3m^2-2m-1+m^2+2m+1}{(m+1)^2}$
$=\dfrac{4m^2}{(m+1)^2} \ge 0$
$\Rightarrow xy \ge -1$
Dấu "=" xảy ra khi $m=0$
Vậy m=0 thì HPT có nghiệm duy nhất và $min_{xy}=-1$

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 )

`x+my=m+1=>x=m+1-my` thế vào dưới

`=>m(m+1-my)+y-3m+1=0`

`<=>m^2+m-my^2+y-3m-1`

`=>y(1-m^2)=2m-1-m^2`

Hệ có no duy nhất

`=>1-m^2 ne 0=>m ne +-1`

`=>y=(-1+2m-m^2)/(1-m^2)=(m-1)/(m+1)`

`=>x=m+1-my=((m+1)^2-m(m-1))/(m+1)=(3m+1)/(m+1)`

`=>xy=((3m+1)(m-1))/(m+1)^2=(3m^2-2m-1)/(m+1)^2`

Xét `xy+1`

`=(3m^2-2m-1+m^2+2m+1)/(m+1)^2=(4m^2)/(m+1)^2`

`=>xy+1>=0=>xy>=-1`

Dấu "=" xảy ra khi `m=0`

giúp mình vớiii

thay m=2 vào HPT ta có
\(\left\{{}\begin{matrix}x+2y=2+1\\2x+y=2.2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=3\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=6\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=2\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
vậy ..........
 

(x:y)=(2;3)

\(\Leftrightarrow\left\{{}\begin{matrix}2-3m=0\\2m-3=m+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2-3m=0\\m-4=0\end{matrix}\right.\)

\(\Leftrightarrow2-3m=m-4\)

\(\Leftrightarrow4m=6\)

\(\Leftrightarrow m=\dfrac{3}{2}\)

Thay x=2 và y=3 vào HPT, ta được:

\(\left\{{}\begin{matrix}2-3m=0\\2m-3=m+1\end{matrix}\right.\Leftrightarrow m\in\varnothing\)

Đề là \(\left(x+y\right)\left(m^2+3\right)=-8\) đúng không?

\(HPT\Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\3x+m\left(mx-2\right)=3m\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow3x+m^2x-2m=3m\\ \Leftrightarrow x\left(m^2+3\right)=5m\Leftrightarrow x=\dfrac{5m}{m^2+3}\\ \Leftrightarrow y=mx-2=\dfrac{5m^2}{m^2+3}-2=\dfrac{3m^2-6}{m^2+3}\\ \Leftrightarrow x+y=\dfrac{5m+3m^2-6}{m^2+3}\\ \left(x+y\right)\left(m^2+3\right)=-8\\ \Leftrightarrow3m^2+5m-6=-8\\ \Leftrightarrow3m^2+5m+2=0\\ \Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{2}{3}\\m=-1\end{matrix}\right.\)