If the sum of two opposite numbers is zero, then:

(x+y- 1)^2 + √(2x-y+4) = 0

∵(x+y- 1)^2≥0,√(2x-y+4)≥0

∴(x+y- 1)^2 = 0，

√(2x-y+4) = 0

Simultaneous: x=- 1, y=2, then:

x^2 + √(y^3) = 1+2√2

2. Solution:

∫4 = 8-√ 16 & lt； 8-√ 1 1 & lt； 8-√9 = 5

∴x= 4

y = 8-√ 1 1-4 = 4-√ 1 1

Then:

2xy-y^2=y(2x-y)=(4-√ 1 1)(4+√ 1 1)= 16- 1 1=5