$1\le\frac{1}{1-x^2}$
$-2^5+4^3-6^1$
$\left(-2\right)+8\left(-5\right)+\left(10\right)$
$2x^2-x-15$
$0-\left(-2\right)$
$3\left(-1\right)^3+2\left(-1\right)^2-\left(-1\right)+7$
$7\left(a+b-c\right);\:a=5;\:b=-4;c=-1$
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