Difference between revisions of "2005 AMC 10A Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | <math> ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \ | + | <math> ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{\textbf{(C) }}</math> |
==Video Solution== | ==Video Solution== | ||
Revision as of 10:30, 7 December 2021
Contents
Problem
For each pair of real numbers 
, define the operation 
as
.
What is the value of 
?
Solution
Video Solution
CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
