Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 8"

Revision as of 13:43, 6 May 2007

If we subtract from $2$ the inverse number of $x-1$ , we get the inverse of $x-1$ . Then the number $x+1$ equals to

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } -1\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } \frac{1}{2}$

$2-\frac1{x-1}=\frac1{x-1}$

$2=\frac2{x-1}$

$2x-2=2$

$2x=4$

$x=2$

$x+1=3\Rightarrow\mathrm{ D}$

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 ==Problem==
-If we substract from <math>2</math> the inverse number of <math>x-1</math>, we get the inverse of <math>x-1</math>. Then the number <math>x+1</math> equals to
+If we subtract from <math>2</math> the inverse number of <math>x-1</math>, we get the inverse of <math>x-1</math>. Then the number <math>x+1</math> equals to
 <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } -1\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } \frac{1}{2}</math>