Difference between revisions of "2006 AMC 10B Problems/Problem 7"

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<math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} =  \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|</math>
 
<math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} =  \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|</math>
  
Since <math>x<0</math>
+
Since <math>x<0,|x|= \boxed{\textbf{(A)}-x} </math>
 
 
<math>|x|= -x \Rightarrow A </math>
 
  
 
== See Also ==
 
== See Also ==

Revision as of 13:46, 26 January 2022

Problem

Which of the following is equivalent to $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$?

$\mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1}$

Solution

$\sqrt{\frac{x}{1-\frac{x-1}{x}}} =  \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|$

Since $x<0,|x|= \boxed{\textbf{(A)}-x}$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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