Question

问题1：已知³√1-2X与³√3Y-2互为相反数.求1+2x/y的值

问题2：在△ABC中，AB=AC，点P在BC上，试说明 AB²=PA²+PB×PC ？

Answer 1

1.³√(1-2X)与³√(3Y-2)互为相反数，
∴1-2x=-(3y-2),
∴3y=1+2x,
∴(1+2x)/y=3.

2.作AD⊥BC于D,则BD=DC.
∴PA²+PB×PC
=PA^+(BD-PD)(BD+PD)
=PA^+BD^-PD^
=BD^+AD^
=AB^.

Answer 2

1.³√(1-2X)与³√(3Y-2)互为相反数，
∴1-2x=-(3y-2),
∴3y=1+2x,
∴(1+2x)/y=3.

2.作AD⊥BC于D,则BD=DC.
∴PA²+PB×PC
=PA^+(BD-PD)(BD+PD)
=PA^+BD^-PD^
=BD^+AD^
=AB^望采纳O(∩_∩)O