(1)3−2−2cos45°+(6−3)0 (2)先化简,再求值:2a(a+b)-(a+b)2,其中:a=2011,b=2010.

问题描述:

(1)3−2−2cos45°+(

6
−3)0
(2)先化简,再求值:2a(a+b)-(a+b)2,其中:a=
2011
b=
2010

(1)原式=

1
9
-2×
2
2
+1
=
1
9
-
2
+1
=
10
9
-
2

(2)原式=2a(a+b)-(a+b)2
=(a+b)(2a-a-b)
=(a+b)(a-b)
=a2-b2
a=
2011
b=
2010
时,原式=(
2011
2+(
2010
2=2011+2010=4021.