题目内容
已知公式
=
+
(R1≠R2),则表示R1的公式是( )
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
A、R1=
| ||
B、R1=
| ||
C、R1=
| ||
D、R1=
|
分析:先把公式变形为
=
-
,然后通分,再转化为求
的倒数即可.
| 1 |
| R1 |
| 1 |
| R |
| 1 |
| R2 |
| 1 |
| R1 |
解答:解:∵
=
+
,
∴
=
-
,即
=
,
∴R1=
.
故选D.
| 1 |
| R |
| 1 |
| R1 |
| 1 |
| R2 |
∴
| 1 |
| R1 |
| 1 |
| R |
| 1 |
| R2 |
| 1 |
| R1 |
| R2-R |
| RR2 |
∴R1=
| RR2 |
| R2-R |
故选D.
点评:本题主要考查异分式的减法运算,还涉及到倒数问题.分式的加减运算中如果是异分母分式,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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