问题
解答题
计算: (1)(
(2)(
(3)(
(4)(
|
答案
(1)原式=x+1;
(2)∵a2≥0,∴原式=a2;
(3)∵a2+2a+1=(a+1)2≥0,
∴原式=a2+2a+1;
(4)∵4x2-12x+9=(2x-3)2≥0,
∴原式=4x2-12x+9.
计算: (1)(
(2)(
(3)(
(4)(
|
(1)原式=x+1;
(2)∵a2≥0,∴原式=a2;
(3)∵a2+2a+1=(a+1)2≥0,
∴原式=a2+2a+1;
(4)∵4x2-12x+9=(2x-3)2≥0,
∴原式=4x2-12x+9.