问题
解答题
计算(1)(
(2)-24+
(3)先化简,再求值:(3x2y-xy2)-3(x2y-2xy2),其中x=
(4)解方程:
(5)解方程组
|
答案
(1)(
-1 3
+1 6
)×(-24),1 4
=
×(-24)-1 3
×(-24)+1 6
×(-24),1 4
=-8+4-6,
=-10;
(2)原式=-24+
×(6+16),1 2
=-24+
×22,1 2
=-16+11,
=-5;
(3)(3x2y-xy2)-3(x2y-2xy2),
=3x2y-xy2-3x2y+6xy2,
=5xy2,
当x=
,y=-1 2
时,1 3
原式=5×
×(-1 2
)2,1 3
=5×
×1 2 1 9
=
;5 18
(4)去分母得,4(2x-1)-3(3x-1)=24,
去括号得,8x-4-9x+3=24,
移项、合并同类项得,-x=25,
系数化为1得,x=-25;
(5)
,6x-3y=-3① 5x-9y=4②
①×3-②得,13x=-13,
解得x=-1,
把x=-1代入①得,-6-3y=-3,
解得y=-1,
故原方程组的解为
.x=-1 y=-1