题目内容
先化简,再求值:
(1)(4a2-3a)-(2a2+a-1)+(2-a2+4a),其中a=2.
(2)已知A=3a2-5ab-b2,B=-2a2+3ab+b2,其中a=-1,b=
,求-3(A-B)+2(B+3A)的值.
(1)(4a2-3a)-(2a2+a-1)+(2-a2+4a),其中a=2.
(2)已知A=3a2-5ab-b2,B=-2a2+3ab+b2,其中a=-1,b=
| 2 |
| 3 |
(1)原式=4a2-3a-2a2-a+1+2-a2+4a
=a2+3,
当a=2时,原式=22+3=7;
(2)原式=-3A+3B+2B+6A
=3A+5B,
∵A=3a2-5ab-b2,B=-2a2+3ab+b2,
∴原式=3(3a2-5ab-b2)+5(-2a2+3ab+b2)
=9a2-15ab-3b2-10a2+15ab+5b2,
=-a2+2b2,
当a=-1,b=
时,原式=-(-1)2+2×(
)2=
.
=a2+3,
当a=2时,原式=22+3=7;
(2)原式=-3A+3B+2B+6A
=3A+5B,
∵A=3a2-5ab-b2,B=-2a2+3ab+b2,
∴原式=3(3a2-5ab-b2)+5(-2a2+3ab+b2)
=9a2-15ab-3b2-10a2+15ab+5b2,
=-a2+2b2,
当a=-1,b=
| 2 |
| 3 |
| 2 |
| 3 |
| 1 |
| 9 |
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