2021 AMC 12A Problems/Problem 7

Revision as of 13:53, 11 February 2021 by Aop2014 (talk | contribs) (Solution)

Problem

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the trivial inequality the minimum value for this is $1$, which can be achieved at $x=y=0$. ~aop2014

Note

See problem 1.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 12 Problems and Solutions

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