Question

已知x的4次方（y的2次方-1）+2x的2次方（y的2次方-1）+（y的2次方-1)=0,利用因式分解求y的值.

Answer 1

x^4(y^2-1)+2x^2(y^2-1)+(Y^2-1)

=(Y^2-1)(x^4+2x^2+1)

=(y-1)(y+1)(x^2+1)^2=0

y-1=0 y=1

y+1=0 y=-1