题目内容
计算:
+
+
+
.
| 1 |
| 1-x |
| 1 |
| 1+x |
| 2 |
| 1+x2 |
| 4 |
| 1+x4 |
分析:首先把前两个分式通分相加,然后依次计算即可求解.
解答:解:原式=
+
+
+
=
+
+
=
+
+
=
+
=
+
=
.
| 1+x |
| (1-x)(1+x) |
| 1-x |
| (1-x)(1+x) |
| 2 |
| 1+x2 |
| 4 |
| 1+x4 |
=
| 2 |
| (1-x)(1+x) |
| 2 |
| 1+x2 |
| 4 |
| 1+x4 |
=
| 2(1+x2) |
| (1-x2)(1+x2) |
| 2(1-x2) |
| (1-x2)(1+x2) |
| 4 |
| 1+x4 |
=
| 4 |
| 1-x4 |
| 4 |
| 1+x4 |
=
| 4(1+x4) |
| (1-x4)(1+x4) |
| 4(1+x4) |
| (1-x4)(1+x4) |
=
| 8 |
| 1-x8 |
点评:本题考查了分式的加减运算,分式的加减运算中,如果是同分母分式,那么分母不变,把分子直接相加减即可;如果是异分母分式,则必须先通分,把异分母分式化为同分母分式,然后再相加减.
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