1: Để hệ có nghiệm duy nhất thì \(\frac14<>\frac{m+1}{-1}\)
=>m+1<>-4
=>m<>-5
\(\begin{cases}x+\left(m+1\right)y=1\\ 4x-y=-2\end{cases}\Rightarrow\begin{cases}4x+\left(4m+4\right)y=4\\ 4x-y=-2\end{cases}\)
=>\(\begin{cases}4x+\left(4m+4\right)y-4x+y=4+2=6\\ 4x-y=-2\end{cases}\Rightarrow\begin{cases}y\left(4m+5\right)=6\\ 4x=y-2\end{cases}\)
=>\(\begin{cases}y=\frac{6}{4m+5}\\ 4x=\frac{6}{4m+5}-2=\frac{6-8m-10}{4m+5}=\frac{-8m-4}{4m+5}\end{cases}\)
=>\(\begin{cases}y=\frac{6}{4m+5}\\ x=\frac{-2m-1}{4m+5}\end{cases}\)
Để (x;y) nguyên thì \(\begin{cases}6\vdots4m+5\\ -2m-1\vdots4m+5\end{cases}\Rightarrow\begin{cases}6\vdots4m+5\\ -4m-2\vdots4m+5\end{cases}\)
=>\(\begin{cases}6\vdots4m+5\\ -4m-5+3\vdots4m+5\end{cases}\Rightarrow\begin{cases}6\vdots4m+5\\ 3\vdots4m+5\end{cases}\Rightarrow3\vdots4m+5\)
=>4m+5∈{1;-1;3;-3}
=>4m∈{-4;-6;-2;-8}
=>m∈{-1;-3/2;-1/2;-2}
mà m nguyên và m<>-5
nên m∈{-1;-2}
2: \(x^2+y^2=0,25\)
=>\(\left(\frac{6}{4m+5}\right)^2+\left(\frac{-2m-1}{4m+5}\right)^2=0,25=\frac14\)
=>\(\frac{36}{\left(4m+5\right)^2}+\frac{\left(2m+1\right)^2}{\left(4m+5\right)^2}=\frac14\)
=>\(\left(4m+5\right)^2=4\left\lbrack\left(2m+1\right)^2+36\right\rbrack\)
=>\(16m^2+40m+25=4\cdot\left\lbrack4m^2+4m+1+36\right\rbrack=4\left(4m^2+4m+37\right)\)
=>\(16m^2+40m+25=16m^2+16m+148\)
=>24m=123
=>\(m=\frac{123}{24}=\frac{41}{8}\) (nhận)
