问题
解答题
(1)计算:20120-3tan30°+(
(2)先化简,再求值:a(a-2b)+2(a+b)(a-b)+(a+b)2,其中a=-
|
答案
(1)原式=1-3×
+9-(2-3 3
)=1-3
+9-2+3
=8;3
(2)原式=a2-2ab+2a2-2b2+a2+2ab+b2=4a2-b2,
当a=-
,b=1时,原式=4×(-1 2
)2-12=1-1=0.1 2
(1)计算:20120-3tan30°+(
(2)先化简,再求值:a(a-2b)+2(a+b)(a-b)+(a+b)2,其中a=-
|
(1)原式=1-3×
+9-(2-3 3
)=1-3
+9-2+3
=8;3
(2)原式=a2-2ab+2a2-2b2+a2+2ab+b2=4a2-b2,
当a=-
,b=1时,原式=4×(-1 2
)2-12=1-1=0.1 2