(1)计算:(π+3)0-|-2013|+
(2)已知a2+2a=-1,求2a(a+1)-(a+2)(a-2)的值. |
(1)原式=1-2013+8×
=1-2013+1=-2011; 1 8
(2)原式=2a2+2a-a2+4=a2+2a+4,
∵a2+2a=-1,
∴原式=-1+4=3.
(1)计算:(π+3)0-|-2013|+
(2)已知a2+2a=-1,求2a(a+1)-(a+2)(a-2)的值. |
(1)原式=1-2013+8×
=1-2013+1=-2011; 1 8
(2)原式=2a2+2a-a2+4=a2+2a+4,
∵a2+2a=-1,
∴原式=-1+4=3.