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 )

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{m}{3}\\-2y=\dfrac{10}{3}m+4\end{matrix}\right.\)

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

Để \(x^2+y^2=10\)

\(\Leftrightarrow\left(\dfrac{-m}{3}\right)^2+\left(\dfrac{-5x}{3}-2\right)^2=10\)

\(\Leftrightarrow\dfrac{m^2}{9}+\dfrac{25m^2}{9}+\dfrac{20m}{3}+4=10\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{20m}{3}-6=0\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{60m}{9}-\dfrac{54}{9}=0\)

\(\Leftrightarrow26m^2+60m-54=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=\dfrac{9}{13}\end{matrix}\right.\)

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

Vì \(\frac31<>\frac{-2}{-1}\left(3<>2\right)\)

nên hệ luôn có nghiệm duy nhất

\(\begin{cases}3x-2y=2m^2-3\\ x-y=3\end{cases}\Rightarrow\begin{cases}3x-2y=2m^2-3\\ 2x-2y=6\end{cases}\)

=>\(\begin{cases}3x-2y-2x+2y=2m^2-3-6\\ x-y=3\end{cases}\Rightarrow\begin{cases}x=2m^2-9\\ y=x-3=2m^2-9-3=2m^2-12\end{cases}\)

\(x^2-y^2=-15\)

=>(x-y)(x+y)=-15

=>\(\left(2m^2-9-2m^2+12\right)\left(2m^2-9+2m^2-12\right)=-15\)

=>\(3\left(4m^2-21\right)=-15\)

=>\(4m^2-21=-5\)

=>\(4m^2=16\)

=>\(m^2=4\)

=>m=2 hoặc m=-2

Để hệ có nghiệm duy nhất thì \(\frac11<>\frac{-1}{-3}\) (luôn đúng)

=>Hệ luôn có nghiệm duy nhất

\(\begin{cases}x-y=4m+8\\ x-3y=6-2m^2\end{cases}\Rightarrow\begin{cases}x-y-x+3y=4m+8-6+2m^2\\ x-y=4m+8\end{cases}\)

=>\(\begin{cases}2y=2m^2+4m+2\\ x=y+4m+8\end{cases}\Rightarrow\begin{cases}y=m^2+2m+1=\left(m+1\right)^2\\ x=m^2+2m+1+4m+8=m^2+6m+9=\left(m+3\right)^2\end{cases}\)

\(\sqrt{x}+\sqrt{y}=8\)

=>\(\sqrt{\left(m+1\right)^2}+\sqrt{\left(m+3\right)^2}=8\)

=>|m+1|+|m+3|=8

=>m+1+m+3=8(Do m>0)

=>2m+4=8

=>2m=4

=>m=2(nhận)

(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:

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

Vì 1/1<>2/1

nên hệ luôn có nghiệm duy nhất

\(\begin{cases}x+2y=5-m\\ x+y=1\end{cases}\Rightarrow\begin{cases}x+2y-x-y=5-m-1\\ x+y=1\end{cases}\)

=>\(\begin{cases}y=4-m\\ x=1-\left(4-m\right)=1-4+m=m-3\end{cases}\)

2x+y<3-4m

=>2(m-3)+4-m<3-4m

=>2m-6+4-m<3-4m

=>m-2<3-4m

=>5m<5

=>m<1

Để hệ có nghiệm duy nhất thì \(\frac{m^2}{m+1}<>\frac{2}{-1}=-2\)

=>\(m^2<>-2m-2\)

=>\(m^2+2m+2<>0\)

=>\(\left(m+1\right)^2+1<>0\) (luôn đúng)

=>Hệ luôn có nghiệm duy nhất

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

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

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

x>0 và y<0

=>\(\begin{cases}\frac{m}{m^2+2m+2}<0\\ \frac{m+2}{m^2+2m+2}>0\end{cases}\Rightarrow\begin{cases}m<0\\ m+2>0\end{cases}\Rightarrow-2