Question

已知:y=根号下1-4x 根号下4x-1+1/2,求代数式根号下x^2+2xy+y^2-根号下x

2-2xy+y^2的值

Answer 1

∵
y=根号下（1-4x）+ 根号下（4x-1）+1/2
1-4x≥0 4x-1≥0
∴
x=1/4
y=1/2
∴
x²+2xy+y²-根号(x²-2xy+y²)
=(x+y)²-根号(x-y)²
=x+y-y+x
=2x
=2*1/4
=1/2