当a=12,b=8时,下列各式的值是多少?
a2+b2=2(a+b)=4a=4b=ab=a2-b2=4a+4b=a2+4a=b2-4b=2(a+b)+ab=
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解:当a=12,b=8时;
(1)a2+b2=122+82=208;
(2)4a=12×4=48;
(3)ab=12×8=96;
(4)4a+4b,
=4×(12+8),
=80;
(5)b2-4b,
=82-8×4,
=32;
(6)2(a+b),
=2×(12+8),
=40;
(7)4b=4×8=32,;
(8)a2-b2=122-82=144-64=80;
(9)a2+4a=122+4×12=192;
(10)2(a+b)+ab,
=2×(12+8)+12×8,
=40+96,
=136.
故