(1)(
a2
a−2+[4/2−a])•[1
a2+2a
=(
a2/a−2]-[4/a−2])•[1
a(a+2)
=
(a+2)(a−2)/a−2]•[1
a(a+2)
=
1/a],
当a=1时,原式=1;
(2)(x2+y2)2-(x2+y2)-12=0,
设x2+y2=a,则有a2-a-12=0,
因式分解得:(a-4)(a+3)=0,
解得:a1=4,a2=-3,
∵x2+y2>0,即a>0,
∴a=-3不合题意,舍去,
则x2+y2=a=4.
(1)(
a2
a−2+[4/2−a])•[1
a2+2a
=(
a2/a−2]-[4/a−2])•[1
a(a+2)
=
(a+2)(a−2)/a−2]•[1
a(a+2)
=
1/a],
当a=1时,原式=1;
(2)(x2+y2)2-(x2+y2)-12=0,
设x2+y2=a,则有a2-a-12=0,
因式分解得:(a-4)(a+3)=0,
解得:a1=4,a2=-3,
∵x2+y2>0,即a>0,
∴a=-3不合题意,舍去,
则x2+y2=a=4.