$-1^4\:>\left(-1\right)^4$
$\left(4+x\right)y=y^3$
$\left(6a+2b-3c\right)\left(-5a^2-2b^2-6c^2\right)$
$\int\left(4-x^2\right)^{\left(\frac{3}{2}\right)}dx$
$\int\frac{2x^3+x^2-2x-7}{x^2-x-2}dx$
$x^2-2x+y^2-4y+\frac{11}{4}=0$
$7=\frac{v}{9}$
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