計(jì)算: (1)x2-(x3y-2x2y2)﹢xy (2)(1-1/2)×(1+1/2)×(1+1/4)×(1+1/16) (3)(2x-3y)2-(y+3x)(3x-y) (4)(2a3b2-4a4b3+6a5b4)÷(-2a3b2)
計(jì)算:
(1)x
2-(x
3y-2x
2y
2)﹢xy
(2)(1-
)×(1+
)×(1+
)×(1+
)
(3)(2x-3y)
2-(y+3x)(3x-y)
(4)(2a
3b
2-4a
4b
3+6a
5b
4)÷(-2a
3b
2)
優(yōu)質(zhì)解答
(1)原式=x
2-x
3y+2x
2y
2+xy;
(2)原式=(1-
)×(1+
)×(1+
)=(1-
)(1+
)=1-
=
;
(3)原式=4x
2-12xy+9y
2-9x
2+y
2=-5x
2-12xy+10y
2;
(4)原式=-1+2ab-3a
2b
2.
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