＜訂正版＞

(1) \(m^{2}\)\(x_{0}\) + m\(y_{0}\) = \(m^{2}\)
+) \(x_{0}\) - m\(y_{0}\) = m + 1
-------------------------
(\(m^{2}\) + 1)\(x_{0}\) = \(m^{2}\) + m + 1.

　m\(x_{0}\) + \(y_{0}\) = m
-) m\(x_{0}\) - \(m^{2}\)\(y_{0}\) = \(m^{2}\) + m
-------------------------
(\(m^{2}\) + 1)\(y_{0}\) = -\(m^{2}\)

故に
\(z_{0}\) = \(x_{0}\) - \(y_{0}\) = (2\(m^{2}\) + m + 1)/(\(m^{2}\) + 1). (前回、足算を間違えた)

(2) 2\(m^{2}\) + m + 1 = 2(\(m^{2}\) + \(\frac{m}{2}\)) + 1 = 2(m + \(\frac{1}{4}\)\()^{2}\) + \(\frac{7}{8}\) ≧ \(\frac{7}{8}\) > 0.
だから \(z_{0}\) = \(x_{0}\) - \(y_{0}\) > 0.
即ち \(x_{0}\) > \(y_{0}\).

(3) \(z_{0}\) = \(x_{0}\) - \(y_{0}\) = 1 だから
2\(m^{2}\) + m + 1 = \(m^{2}\) + 1
即ち \(m^{2}\) + m = m(m + 1) = 0
だから m = 0, -1..

(1) (2\(m^{2}\)+m+1)/(\(m^{2}\)+1)
(2) m の値に関わらず、\(x_{0}\)＞\(y_{0}\) (これは合ってる。)
(3) -1, 0