几道初一的数学题~~~~急~~~~~~有分~~~~~今晚就给~~~~~~本人在线等~~~~~
问题描述：
几道初一的数学题~~~~急~~~~~~有分~~~~~今晚就给~~~~~~本人在线等~~~~~
1、当k是多少时,多项式x^2-3kxy-3y^2+(1/3)xy-8中不含xy的项?
2、已知m^2+n^2-6m+10n+34=0,求m+n的值.
(两道题我都要解题过程哦~~拜托啦~~)
答
1.x^2-3kxy+1/3xy-8 -3y^2
=x^2-(3k-1/3)xy-8 -3y^2
不含xy
所以3k-1/3=0
k=1/9
2.上式=m^2-6m+9+n^2+10+25
=(m-3)^2+(n+5)^2
=0
所以m-3=0 n+5=0
m=3 n=-5
m+n=-2