Tuesday, September 8, 2015

Floor function challenge

Solve the equation $\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{a^2+ 2a +2}{a^2+ 1}\right\}=\dfrac{2a -a^2}{a^2 + 1}$, where…


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