Question

1.根号下(x+2)²+y² 乘 根号下(x-2)²+y² 等于 x²

Answer 1

根号下[(x+2)²+y²]×根号下[(x-2)²+y²]=x²+y²

根号下[(x+2)²y²+(x-2)²y²+(x+2)²(x-2)²+(y²)²]=x²+y²

[ (x²)²-8x²+16+2x²y²+8y²+(y²)²]=(x²)²+2x²y²+(y²)²

8x²-8y²=16

x²-y²=2

2 y=y1x/x1 y1=x1y/x

y=-y1(x-0.5) y1=-y/(x-0.5)