1.已知：根号下(x+3y+1) + 2x-y-5的绝对值 = 0,求x,y的值

题目

1.已知：根号下(x+3y+1) + 2x-y-5的绝对值 = 0,求x,y的值
2.已知x=y+ 3 根号下3,求x^2 + x +y^2 -2xy -y 的值

答案

根号下(x+3y+1) + 2x-y-5的绝对值 = 0
因为根式和绝对值都是非负数,都>=0
所以:x+3y+1=0,2x-y-5=0
解得:
x=2,y=-1
2:
x=y+ 3 根号下3,
x-y=3根号3
x^2 + x +y^2 -2xy -y
=(x-y)^2+(x-y)
=(x-y)(x-y+1)
=3根号3*(3根号3+1)
=27+3根号3