Question

实数 $x$ 和 $y$ 满足方程 $x+y$ 是多少？ $x^2+y^2=10x-6y-34.$

Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34.$ What is $x+y?$

$1$

$2$

$3$

$6$

$8$

答案：B

知识点：配方法

难度评级：1020

解答：

方程可化为配方得 $x^2-10x+y^2+6y+34=0.$

两个平方项都不小于 $0$ ，而它们的和为 $0$ ，所以两者都等于 $0$ ，即 $x^2-10x+25+y^2+6y+9$ $=(x-5)^2+(y+3)^2$ $=0.$ $(x-5)^2 = (y+3)^2=0.$

因此 $x=5,y=-3$ ，所以 $x+y=2$ 。

所以正确答案是 B。

This can be rewritten as $x^2-10x+y^2+6y+34=0.$

From this, completing the square yields $x^2-10x+25+y^2+6y+9$ $=(x-5)^2+(y+3)^2$ $=0.$ Since both of these squared terms must be greater than or equal to $0,$ and their sum equals $0,$ both values must both be $0$ yielding $(x-5)^2 = (y+3)^2=0.$

As such, $x=5,y=-3.$ This makes the sum $x+y=2.$

Thus, the correct answer is B .

2013 AMC 10B 第 11 题

答案：B

难度评级：1020

解答：

其他年份的第 11 题