问题 解答题
计算(1)(
1
3
-
1
6
+
1
4
)×(-24).
(2)-24+
1
2
×[6+(-4)2]

(3)先化简,再求值:(3x2y-xy2)-3(x2y-2xy2),其中x=
1
2
,y=-
1
3
.

(4)解方程:
2x-1
6
-
3x-1
8
=1

(5)解方程组
6x-3y=-3
5x-9y=4
答案

(1)(

1
3
-
1
6
+
1
4
)×(-24),

=

1
3
×(-24)-
1
6
×(-24)+
1
4
×(-24),

=-8+4-6,

=-10;

(2)原式=-24+

1
2
×(6+16),

=-24+

1
2
×22,

=-16+11,

=-5;

(3)(3x2y-xy2)-3(x2y-2xy2),

=3x2y-xy2-3x2y+6xy2

=5xy2

当x=

1
2
,y=-
1
3
时,

原式=5×

1
2
×(-
1
3
2

=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

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