问题 解答题
计算
(1)-2-2-
3-27
+(π-1)0-|-1+
1
4
|+(
3
+
2
)
2
•(5-2
6
);
(2)已知:a=
2
-1
2
+1
,b=
27
+
6
3
,求
a2+2ab+b2
-
a2-2ab+b2

(3)解方程组:
1-0.3(y-2)=
x+1
5
y-3
4
=
4x+9
20
-1.5

(4)解不等式组:
2(x+8)≤10-4(x-3)
x+1
3
-
3x+1
2
<1
(并把解集在数轴上表示出来).
答案

(1)原式=-

1
4
-(-3)+1-
3
4
+(5+2
6
)(5-2
6

=-

1
4
+3+1-
3
4
+25-24

=4;

(2)∵a=

2
-1
2
+1
=3-2
2
,b=
27
+
6
3
=3+
2

∴原式=

(a+b)2
-
(a-b)2

=|a+b|-|a-b|

=6-

2
-3
2

=6-4

2

(3)

1-0.3(y-2)=
x+1
5
y-3
4
=
4x+9
20
-1.5②

由①去分母得:10-3(y-2)=2(x+1),

去括号得:10-3y+6=2x+2,即2x+3y=14③,

由②去分母得:5(y-3)=4x+9-30,

去括号得:5y-15=4x-21,即4x-5y=6④,

③×2-④得:11y=22,

解得:y=2,

把y=2代入③得:2x+6=14,

解得:x=4,

x=4
y=2

(4)

2(x+8)≤10-4(x-3)①
x+1
3
-
3x+1
2
<1②

由①去括号得:2x+16≤10-4x+12,即6x≤6,

解得:x≤1,

由②去分母得:2(x+1)-3(3x+1)<6,

去括号得:2x+2-9x-3<6,即-7x<7,

解得:x>-1,

则原不等式的解集为-1<x≤1,表示在数轴上,如图所示:

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