Problem: For xβ 0,1x+12x+13xx \neq 0, \dfrac{1}{x}+\dfrac{1}{2 x}+\dfrac{1}{3 x}xξ =0,x1β+2x1β+3x1β equals
Answer Choices:
A. 12x\dfrac{1}{2 x}2x1β
B. 16x\dfrac{1}{6 x}6x1β
C. 56x\dfrac{5}{6 x}6x5β
D. 116x\dfrac{11}{6 x}6x11β
E. 16x3\dfrac{1}{6 x^{3}}6x31β
Solution:
1x+12x+13x=66x+36x+26x=116x\dfrac{1}{x}+\dfrac{1}{2 x}+\dfrac{1}{3 x}=\dfrac{6}{6 x}+\dfrac{3}{6 x}+\dfrac{2}{6 x}=\dfrac{11}{6 x}x1β+2x1β+3x1β=6x6β+6x3β+6x2β=6x11β.