Để hệ có nghiệm duy nhất thì \(\frac{m+1}{m}<>\frac{-1}{1}\)

=>m+1<>-m

=>2m+1<>0

=>m<>-1/2

\(\begin{cases}\left(m+1\right)x-y=3\\ mx+y=m\end{cases}\Rightarrow\begin{cases}\left(m+1\right)x-y+mx+y=m+3\\ mx+y=m\end{cases}\)

=>\(\begin{cases}x\left(2m+1\right)=m+3\\ y=m-mx\end{cases}\Rightarrow\begin{cases}x=\frac{m+3}{2m+1}\\ y=m-m\cdot\frac{m+3}{2m+1}=\frac{2m^2+m-m^2-3m}{2m+1}=\frac{m^2-2m}{2m+1}\end{cases}\)

x+y>0

=>\(\frac{m^2-2m+m+3}{2m+1}>0\)

=>\(\frac{m^2-m+3}{2m+1}>0\)

=>2m+1>0

=>2m>-1

=>m>-1/2

Để hệ có nghiệm duy nhất thì \(\frac{m-1}{m}<>\frac11\)

=>m-1<>m

=>-1<>0(luôn đúng)

=>Hệ luôn có nghiệm duy nhất

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

=>\(\begin{cases}-x=1-m\\ y=m+1-mx\end{cases}\Rightarrow\begin{cases}x=m-1\\ y=m+1-m\left(m-1\right)=m+1-m^2+m=-m^2+2m+1\end{cases}\)

2x+y<=3

=>\(2\left(m-1\right)-m^2+2m+1-3\le0\)

=>\(2m-2-m^2+2m-2\le0\)

=>\(-m^2+4m-4\le0\)

=>\(-\left(m-2\right)^2\le0\) (luôn đúng)

=>ĐPCM

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 )

Lời giải:

Khi \(m=-\sqrt{2}\). HPT tương đương:

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

Cộng theo vế: \(\Rightarrow (1-2\sqrt{2})x=3-\sqrt{2}\Rightarrow x=\frac{3-\sqrt{2}}{1-2\sqrt{2}}=\frac{1-5\sqrt{2}}{7}\)

\(\Rightarrow y=(m+1)x-3=\frac{(-\sqrt{2}+1)(1-5\sqrt{2})}{7}-3=-\frac{10+6\sqrt{2}}{7}\)

b)

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

\(\Rightarrow mx+[(m+1)x-3]=m\)

\(\Leftrightarrow x(2m+1)=m+3\)

Để hệ có bộ nghiệm duy nhất thì $x$ là duy nhất.

Với \(m=-\frac{1}{2}\Rightarrow x.0=\frac{5}{2}\) (vô lý, pt vô nghiệm)

Với \(m\neq -\frac{1}{2}\), pt có nghiệm duy nhất \(x=\frac{m+3}{2m+1}\)

\(\Rightarrow y=(m+1)x-3=\frac{m^2-2m}{2m+1}\)

Do đó: \(x+y=\frac{m^2-m+3}{2m+1}\)

Để \(x+y>0\Leftrightarrow \frac{m^2-m+3}{2m+1}>0\Leftrightarrow \frac{(m-\frac{1}{2})^2+\frac{11}{4}}{2m+1}>0\)

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

Vậy đk là \(m> \frac{-1}{2}\)

(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:

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

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 ..........
 

a/ Xét pt : \(\left\{{}\begin{matrix}mx-y=1\\\dfrac{x}{2}-\dfrac{y}{2}=335\end{matrix}\right.\)

 Khi \(m=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-669\\y=-1339\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}mx-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\mx-\left(x-670\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\x\left(m-1\right)=-669\end{matrix}\right.\)

Để pt có nghiệm duy nhất \(\Leftrightarrow m\ne1\)

Vậy...