《2025年云南省标准教辅优佳学案配套测试卷八年级数学上册人教版》

26. (本小题满分10分)
先阅读下面的内容，再解答问题.
例题：若$m^2 + 2mn + 2n^2 - 6n + 9 = 0$，求$m$和$n$的值.
解：$\because m^2 + 2mn + 2n^2 - 6n + 9 = 0$，
$\therefore (m^2 + 2mn + n^2) + (n^2 - 6n + 9) = 0$.
$\therefore (m + n)^2 + (n - 3)^2 = 0$.
$\therefore m + n = 0$，$n - 3 = 0$.
解得$m = -3$，$n = 3$.
问题：
(1)若$x^2 + 2y^2 - 2xy + 6y + 9 = 0$，求$x^y$的值.
(2)已知$\triangle ABC$的三边长$a$，$b$，$c$都是正整数，且满足$a^2 + b^2 - 6a - 4b + 13 + |3 - c| = 0$，那么$\triangle ABC$是什么形状的三角形？
(1)
$\because x^2 + 2y^2 - 2xy + 6y + 9 = 0$，
$\therefore (x^2 - 2xy + y^2) + (y^2 + 6y + 9) = 0$，
$\therefore (x - y)^2 + (y + 3)^2 = 0$，
$\because (x - y)^2 \geq 0$，$(y + 3)^2 \geq 0$，
$\therefore x - y = 0$，$y + 3 = 0$，
解得$y = -3$，$x = y = -3$，
$\therefore x^y = (-3)^{-3} = -\dfrac{1}{27}$。
(2)
$\because a^2 + b^2 - 6a - 4b + 13 + |3 - c| = 0$，
$\therefore (a^2 - 6a + 9) + (b^2 - 4b + 4) + |3 - c| = 0$，
$\therefore (a - 3)^2 + (b - 2)^2 + |3 - c| = 0$，
$\because (a - 3)^2 \geq 0$，$(b - 2)^2 \geq 0$，$|3 - c| \geq 0$，
$\therefore a - 3 = 0$，$b - 2 = 0$，$3 - c = 0$，
解得$a = 3$，$b = 2$，$c = 3$，
$\because a = c = 3$，
$\therefore \triangle ABC$是等腰三角形。

答案：
(1)
$\because x^2 + 2y^2 - 2xy + 6y + 9 = 0$，
$\therefore (x^2 - 2xy + y^2) + (y^2 + 6y + 9) = 0$，
$\therefore (x - y)^2 + (y + 3)^2 = 0$，
$\because (x - y)^2 \geq 0$，$(y + 3)^2 \geq 0$，
$\therefore x - y = 0$，$y + 3 = 0$，
解得$y = -3$，$x = y = -3$，
$\therefore x^y = (-3)^{-3} = -\dfrac{1}{27}$。
(2)
$\because a^2 + b^2 - 6a - 4b + 13 + |3 - c| = 0$，
$\therefore (a^2 - 6a + 9) + (b^2 - 4b + 4) + |3 - c| = 0$，
$\therefore (a - 3)^2 + (b - 2)^2 + |3 - c| = 0$，
$\because (a - 3)^2 \geq 0$，$(b - 2)^2 \geq 0$，$|3 - c| \geq 0$，
$\therefore a - 3 = 0$，$b - 2 = 0$，$3 - c = 0$，
解得$a = 3$，$b = 2$，$c = 3$，
$\because a = c = 3$，
$\therefore \triangle ABC$是等腰三角形。