问题
解答题
计算: (1)(
(2)(π-3)0+(
(3)
(4)(
|
答案
(1)原式=1+3
×2
-4=1+3-4=0;2 2
(2)原式=1+
-2 2
=1;2 2
(3)原式=3
-1-6×3
+2=33 2
-1-33
+2=1;3
(4)原式=2+2
+1-2×2
×3 2
=2+23
+1-3=22
.2
计算: (1)(
(2)(π-3)0+(
(3)
(4)(
|
(1)原式=1+3
×2
-4=1+3-4=0;2 2
(2)原式=1+
-2 2
=1;2 2
(3)原式=3
-1-6×3
+2=33 2
-1-33
+2=1;3
(4)原式=2+2
+1-2×2
×3 2
=2+23
+1-3=22
.2