题目内容
计算:
(1)(-
-
+
)÷
(2)|-
|÷(
-
)-
×(-4)2
(3)求(2y2-x2)-2(x2+2y2)的值,其中x=1,y比x的相反数小1.
(1)(-
3 |
4 |
5 |
9 |
7 |
12 |
1 |
36 |
(2)|-
7 |
9 |
2 |
3 |
1 |
5 |
1 |
3 |
(3)求(2y2-x2)-2(x2+2y2)的值,其中x=1,y比x的相反数小1.
(1)原式=(-
-
+
)×36
=-
×36-
×36+
×36
=-27-20+21=-26.
(2)原式=
÷
-
×16
=
×
-
=-
.
(3)化简,得(2y2-x2)-2(x2+2y2)=-2y2-3x2
由x=1,y比x的相反数小1,得y=-2,
当x=1,y=-2时,-2y2-3x2=-2×(-2)2-3=-11.
3 |
4 |
5 |
9 |
7 |
12 |
=-
3 |
4 |
5 |
9 |
7 |
12 |
=-27-20+21=-26.
(2)原式=
7 |
9 |
7 |
15 |
1 |
3 |
=
7 |
9 |
15 |
7 |
16 |
3 |
=-
11 |
3 |
(3)化简,得(2y2-x2)-2(x2+2y2)=-2y2-3x2
由x=1,y比x的相反数小1,得y=-2,
当x=1,y=-2时,-2y2-3x2=-2×(-2)2-3=-11.
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