题目内容
计算:
(1)(
)2(x≥o);
(2)(
)2;
(3)(
)2;
(4)(
)2.
(1)(
| x+1 |
(2)(
| a2 |
(3)(
| a2+2a+a |
(4)(
| 4x2-12x+9 |
(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.
(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.
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