Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer

1. If $m=\frac{(|2 a+1|+|2 b-1|)^{2}-4(a+b)^{2}}{(2 a+1)(2 b-1)}$ , then the value of $m(m+4)$ is:

Pick one

Solution

- 1. B.

Let $2a+1=x, 2b-1=y$ . Then $m=\frac{(|x|+|y|)^{2}-(x+y)^{2}}{xy}=\frac{2|xy|}{xy}-2$ . Therefore, $m(m+4)=(m+2)^{2}-4=\frac{4x^{2}y^{2}}{x^{2}y^{2}}-4=0$ .

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