Question
If $$\frac{a}{{1 - 2a}}$$ $$+$$ $$\frac{b}{{1 - 2b}}$$ $$+$$ $$\frac{c}{{1 - 2c}}$$ = $$\frac{1}{2}{\text{,}}$$ then the value of $$\frac{1}{{1 - 2a}}$$ $$+$$ $$\frac{1}{{1 - 2b}}$$ $$+$$ $$\frac{1}{{1 - 2c}}$$ is?
Answer: Option D
$$\eqalign{
& \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr
& {\text{Multiply by 2 both side}} \cr
& \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr
& {\text{Adding 3 both side}} \cr} $$
$$ \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + $$ $$\frac{{2c}}{{1 - 2c}} = $$ $$1 + 3$$
$$ \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4$$
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$$\eqalign{
& \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr
& {\text{Multiply by 2 both side}} \cr
& \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr
& {\text{Adding 3 both side}} \cr} $$
$$ \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + $$ $$\frac{{2c}}{{1 - 2c}} = $$ $$1 + 3$$
$$ \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4$$
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