PT \(\Leftrightarrow\dfrac{m\left(1-mx\right)+1+mx}{\left(1+mx\right)\left(1-mx\right)}=\dfrac{1}{\left(1-mx\right)\left(1+mx\right)}\)

\(\Rightarrow m-m^2x+1+mx=1\)

\(\Leftrightarrow x\left(m-m^2\right)+m=0\)

Để phương trình vô nghiệm  \(\Leftrightarrow\left\{{}\begin{matrix}m-m^2=0\\m\ne0\end{matrix}\right.\)

\(\Leftrightarrow m=1\)
Vậy \(m=1\)

\(x^2=\left(y+1\right)^2+12\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y+1\right)=12\)

Do \(x,y\in N\)^* nên \(x-y-1;x+y+1\inƯ\left(12\right)\) và \(x+y+1\ge1+1+1=3\)

TH1: \(x+y+1=12\Rightarrow x-y-1=1\)

\(\Leftrightarrow x=\dfrac{13}{2};y=\dfrac{9}{2}\) (ktm)

TH2:\(x+y+1=6;x-y-1=2\)

\(\Leftrightarrow x=4;y=1\) (thỏa mãn)

TH3: \(x+y+1=4;x-y-1=3\)

\(\Leftrightarrow x=\dfrac{7}{2};y=-\dfrac{1}{2}\) (ktm)

TH4: \(x+y+1=3;x-y-1=4\) (ktm)

Vậy \(x=4;y=1\)