1964 AHSME Problems/Problem 4
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Contents
[hide]Problem
The expression
where and , is equivalent to:
Solution 1
Simplifying before substituting by obtaining a common denominator gives:
Substituting gives:
Solution 2
Substituting and then simplifying, noting that and , gives:
Solution 3
If it's true for all values of and , then it's true for a specific value of and . Letting and gives and . The fraction becomes:
Plugging in into the answer choices gives . Therefore, the answer is
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AHSME Problems and Solutions |
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