Mathematics
Junior High

2番と3番の解き方を教えてください!
早急にお願いします!

Answers

(2)訂正
(x-y)^2+3y^2
=(√5+3-√5+3)^2+3(√5-3)^2
=36+3(5+9-6√5)
=36+42-18√5
=78-18√5

Post A Comment

(2)
(x-2y)^2に代入して
=(√5+3-2√5+6)^2
=(9-√5)^2
=86-18√5

(3)
(y-x)/xy
=(√5-3-√5-3)/(√5+3)(√5-3)
=-6/(5-9)
=6/4
=3/2

Post A Comment
Were you able to resolve your confusion?