$\frac{3x^3+2x^2+3x+7}{x^2+1}$

$\frac{7}{x+1}=\frac{6}{x-5}$

$\frac{\left(x^3+9\right)\cdot5}{12}=15$

$y=\:\sqrt[9]{\frac{x-3}{x+9}}$

$\frac{dx}{dt}=8\left(x^2+1\right)$

$\frac{2\sin^2\left(x\right)-1}{\sin\left(x\right)+\cos\left(x\right)}=-\sin\left(x\right)+\cos\left(x\right)$

$3-4-6+4-1-8-6-8$