(1)(ab-3a2)-2b2-5ab-(a2-2ab)(2)12a2b?5ac?(3a2c?a2b)+(3ac?4a2c)(3)先化简再求值-2y3+(3xy2
(1)(ab-3a2)-2b2-5ab-(a2-2ab)(2)12a2b?5ac?(3a2c?a2b)+(3ac?4a2c)(3)先化简再求值-2y3+(3xy2-x2y...
(1)(ab-3a2)-2b2-5ab-(a2-2ab)(2)12a2b?5ac?(3a2c?a2b)+(3ac?4a2c)(3)先化简再求值-2y3+(3xy2-x2y)-2(xy2-y3),其中|2x-2|+(y+1)2=0(4)有理数a、b、c在数轴上的对应点如图所示,化简代数式:|2a-b|+3|a+b|-|4c-a|
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(1)(ab-3a2)-2b2-5ab-(a2-2ab)
原式=ab-3a2-2b2-5ab-a2+2ab
=-4a2-2b2-2ab;
(2)原式=
a2b?5ac?3a2c+a2b+3ac?4a2c
=
a2b?2ac?7a2c;
(3)原式=xy2-x2y,
∵|2x-2|+(y+1)2=0,
∴2x-2=0,y+1=0
∴x=1,y=-1,
∴原式=1×1-1×(-1)
=2;
(4)根据有理数a、b、c在数轴上的如图所示的对应点知,
a<b<0<c,
∴原式=b-2a-3a-3b-4c+a
=-4a-2b-4c.
原式=ab-3a2-2b2-5ab-a2+2ab
=-4a2-2b2-2ab;
(2)原式=
1 |
2 |
=
3 |
2 |
(3)原式=xy2-x2y,
∵|2x-2|+(y+1)2=0,
∴2x-2=0,y+1=0
∴x=1,y=-1,
∴原式=1×1-1×(-1)
=2;
(4)根据有理数a、b、c在数轴上的如图所示的对应点知,
a<b<0<c,
∴原式=b-2a-3a-3b-4c+a
=-4a-2b-4c.
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