题目内容
计算:
(1)
-x-1
(2)
-
(3)(
-
)÷
(4)(1+
)2•(1
)2.
(1)
x2 |
x-1 |
(2)
x2+2x |
x2-4 |
2 |
x-2 |
(3)(
x |
x+1 |
3x |
x-1 |
x |
x2-1 |
(4)(1+
2b |
a-b |
2b |
a+b |
(1)原式=
-
=
=
.
(2)原式=
-
=
-
=
=1.
(3)原式=
•
=
=-2x-4.
(4)原式=(
)2•(
)2
=
•
=
=
.
x2 |
x-1 |
x+1 |
1 |
=
x2-(x+1)(x-1) |
x-1 |
=
1 |
x-1 |
(2)原式=
x(x+2) |
(x+2)(x-2) |
2 |
x-2 |
=
x |
x-2 |
2 |
x-2 |
=
x-2 |
x-2 |
=1.
(3)原式=
x(x-1)-3x(x+1) |
(x+1)(x-1) |
(x+1)(x-1) |
x |
=
-2x2-4x |
x |
=-2x-4.
(4)原式=(
a-b+2b |
a-b |
a+3b |
a+b |
=
(a+b)2 |
(a-b)2 |
(a+3b)2 |
(a+b)2 |
=
(a+3b)2 |
(a-b)2 |
=
a2+6ab+9b2 |
a2-2ab+b2 |
练习册系列答案
相关题目