$x^2-4x-5>1$
$\frac{2x^2-3}{\left(3-2x\right)\left(1-x\right)^2}$
$\left(x^2-4x+2\right)^2$
$1-\cos=\frac{\sin\left(x\right)}{\csc\left(x\right)+\cot\left(x\right)}$
$\left(-572\right)\left(45\right)\left(-1\right)$
$2x^2-5x+1=0$
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