
(1)计算:8?(π?2)0+2cos45°+4?1(2)先化简.再求值:a(a-2b)+2(a+b)(a-b)+(a+b)2,其中a=-1
(1)计算:8?(π?2)0+2cos45°+4?1(2)先化简.再求值:a(a-2b)+2(a+b)(a-b)+(a+b)2,其中a=-12,b=1....
(1)计算:8?(π?2)0+2cos45°+4?1(2)先化简.再求值:a(a-2b)+2(a+b)(a-b)+(a+b)2,其中a=-12,b=1.
展开
1个回答
展开全部
(1)
?(π?2)0+2cos45°+4?1,
=2
-1+2×
+
,
=3
-
;
(2)a(a-2b)+2(a+b)(a-b)+(a+b)2,
=a2-2ab+2a2-2b2+a2+b2+2ab,
=4a2-b2,
当a=-
,b=1,原式=4a2-b2=4×
-1,
=0.
8 |
=2
2 |
| ||
2 |
1 |
4 |
=3
2 |
3 |
4 |
(2)a(a-2b)+2(a+b)(a-b)+(a+b)2,
=a2-2ab+2a2-2b2+a2+b2+2ab,
=4a2-b2,
当a=-
1 |
2 |
1 |
4 |
=0.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询