2/3（x-5)^2+5*(m的绝对值)
是不是2/3（x-5)^2+5*(m的绝对值)=0?
如果是则(x-5)^2>=0,所以2/3(x-5)^2>=0
|m|〉=0,所以5*|m|>=0
两个大于等于0的数相加等于0,若有一个大于0,则另一个小于0,不成立
所以两个都等于0
所以x-5=0,m=0
x=5
-2a^2b^(y+1)与3a^2b^3是同类项
所以a和b的指数对应相等
所以y+1=3,y=2
0.375x^2y+5m^2x-{-7/16x^2y+[-1/4xy^2+(-3/16x^2y-3.475xy^2)]-6.275xy^2}
=0.375x^2y+5*0^(2*5)-[-7/16x^2y+(-1/4xy^2-3/16x^2y-3.475xy^2)-6.275xy^2]
=0.375x^2y+0-(-7/16x^2y-1/4xy^2-3/16x^2y-3.475xy^2-6.275xy^2)
=0.375x^2y-(-5/8x^2y-10xy^2)
=0.375x^2y+5/8x^2y+10xy^2
=x^2y+10xy^2
=5^2*2+10*5*2^2
=50+200
=250