A) \[\frac{2abc}{ac+bc-ab}\]
B) \[\frac{2abc}{ab-ac+bc}\]
C) \[\frac{2abc}{ab+bc+ca}\]
D) \[\frac{2abc}{ab+ac-bc}\]
Correct Answer: A
Solution :
[a] \[\frac{1}{a}=\frac{1}{x}+\frac{1}{y};\frac{1}{b}=\frac{1}{x}+\frac{1}{z};\frac{1}{c}=\frac{1}{y}+\frac{1}{z}\] \[\Rightarrow \,\,\,\,\frac{1}{b}+\frac{1}{a}-\frac{1}{c}=\frac{2}{x}\] \[\Rightarrow \,\,\,\,x=\frac{2abc}{bc+ac-ab}\]You need to login to perform this action.
You will be redirected in
3 sec