问题
解答题
计算与化简: (1)-1
(2)[30-(
(3)4a2+b2+2ab-4a2-b2-7ab; (4)x-3(1+2x-x2)+2(3x-2-x2). |
答案
(1)原式=-
×3 2
×4 3
×1 5
=-1;5 2
(2)原式=[30-
×36-7 9
×36+5 6
×36]×(-11 12
)1 5
=(30-28-30+33)×(-
)1 5
=5×(-
)1 5
=-1;
(3)原式=(4a2-4a2)+(3b2-4b2)+(2ab-7ab)
=b2-5ab;
(4)原式=x-3-6x+3x2+6x-4-2x2
=x+x2+7.