一、计算:
a2b2abx24x2x1))(1)(; (2)(2 abbaab
(3)aba(a2abb2a);
(5)x23xx2(x32x);
xx6x3x3a2(4)ababa2(ba)
6)(1x31x1x21)x1 (b2aba1b(ab) (7)(2; (8) )2abaababba
x2yx2y22a1ab (9) () (10)()2 2babb42y2xyx
(11) (12) ()2()
x12xxx111x1x1x2111(1),其中x 2.先化简,再求值:
x2x23
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