若实数x,y满足x²+y²-10x-2y+26=0 ,求根号下x²y²的值
人气:464 ℃ 时间:2020-04-15 22:31:28
解答
x²-10x+y²-2y+26=0
(x-5)²=x²-10x+25
(x-5)(x-5)=x²-10x+25
(x-5)²+(y-1)²=0
x-5=0y-1=0
x=5y=1
求根号下x²y²=25这里不会使用到因式分解了吧?不会但是我还是看不懂啊……真的不好意思……真的看不懂……哪里看不懂x²-10x+y²-2y+26=0 (x-5)²=x²-10x+25(x-5)(x-5)=x²-10x+25这些都是怎么来的?老师教的 具体怎样。。。(我打字慢)这是对的 给个分吧没功劳也有苦劳吧 就这样写肯定对
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