Question

已知x=+3, y=-3，求下列各式的值；
(1)x²-2xy+y² ,
(2)x²-y²；

Answer 1

(1)20; (2) 12.

試題分析：先由條件求出：x+y=2，x-y=6；再把結論進行變形，代入求值即可.
試題解析：∵x=+3, y=-3
∴x+y=2，x-y=6
(1)x²-2xy+y²=(x+y)²=(2)2="20,"
(2)x²-y²="(x+y)(x-y)=" 2×6=12.