Question

giúp mình bài 2 ạ

Answer 1

\(2,\left\{{}\begin{matrix}2x-y=m-5\\mx+y=m^2-m+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2x-m+5\left(1\right)\\mx+2x-m+5=m^2-m+1\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow x\left(m+2\right)=m^2-m+1+m-5=m^2-4\)

\(hệ\) \(pt\) \(có\) \(ngo\) \(duy\Leftrightarrow m+2\ne0\Leftrightarrow m\ne-2\)

\(b,\Rightarrow\left\{{}\begin{matrix}y=2.\left(m-2\right)-m+5=m+1\\x=\dfrac{m^2-4}{m+2}=m-2\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{m-2}{m+1}=1+\dfrac{-3}{m+1}\in Z\Leftrightarrow\left\{{}\begin{matrix}m\ne-1,m\ne-2\left(để-hệ-pt-có-nghiệm-duy-nhất\right)\\m+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\end{matrix}\right.\)

\(\Rightarrow m=\left\{0;2;-4\right\}\)