Question

若且 $x \neq y$ ，求 $y+4 = (x-2)^2,$ $x+4 = (y-2)^2,$ $x^2+y^2?$

If we have that $y+4 = (x-2)^2,$ $x+4 = (y-2)^2,$ and $x \neq y,$ what is the value of $x^2+y^2?$

$10$

$15$

$20$

$25$

$30$

答案：B

知识点：方程组对称性（代数）代数变形

难度评级：1420

解答：

展开并相加两个方程，得到所以两式相减得到即因为 $x \neq y$ ，可除以 $x - y$ ，得到进而 $x^2 + y^2 - 4x - 4y + 8$ $= x + y + 8.$ $x^2 + y^2 = 5(x + y).$ $x^2 - y^2 - 4x + 4y = y - x.$ $x^2 - y^2 = 3(x - y).$ $x + y = 3$ $x^2 + y^2 = 15.$

所以正确答案是 B。

Adding the two equations gives us $x^2 + y^2 - 4x - 4y + 8$ $= x + y + 8.$ We can rearrange this equation to get $x^2 + y^2 = 5(x + y).$ We can then subtract them to get $x^2 - y^2 - 4x + 4y = y - x.$ Once again rearranging, we can find $x^2 - y^2 = 3(x - y).$ We have that $x \neq y,$ which means that we can divide both sides by $x - y.$ This gives us $x + y = 3$ $x^2 + y^2 = 15.$

Thus, B is the correct answer.

2015 AMC 10A 第 16 题

答案：B

难度评级：1420

解答：

其他年份的第 16 题