已知-7≤x<1,試化簡:根號x²;+14x+49-根號(1-x)²;/1-x.

已知-7≤x<1,試化簡:根號x²;+14x+49-根號(1-x)²;/1-x.

-7≤x0,
∴√(x²;+14x+49)-[√(1-x)²;]/(1-x)
=√(x+7)²;-|1-x|/(1-x)
=(x+7)-(1-x)/(1-x)
=x+7-1
=x+6.