1.若多项式3x^2+1/3xy-3kxy+3y^2+5中不含xy的项,则k的值是?

题目

1.若多项式3x^2+1/3xy-3kxy+3y^2+5中不含xy的项,则k的值是?
还有一题!已知m是绝对是最小的有理数,并且单项式-2a^(m+2)b^y+1与4a^xb^3的和认为单项式,请求出多项式2x^2-3xy+6y^2-(3+m)x^2+mxy-9my^2的值

答案

3x^2+1/3xy-3kxy+3y^2+5
=3x^2+(1/3-3k)xy+3y^2+5
不含xy的项,xy项的系数=0
1/3-3k=0
k=1/9
m是绝对值最小的有理数,m=0
[-2a^(m+2)b^y]+1+4a^xb^3
=-2a^2b^y+4a^xb^3
要结果仍为单项式,则x=2 y=3
2x^2-3xy+6y^2-(3+m)x^2+mxy-9my^2
=2x^2-3xy+6y^2-3x^2
=-x^2-3xy+6y^2
=-4-18+54
=32