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