$\frac{2x^2+4x+6}{\left(2x+3\right)\left(2x+2\right)}=\frac{17}{35}$

$9\left(\frac{2}{3}\right)^2-12\left(-\frac{1}{2}\right)^3-6\left(\frac{2}{3}\right)\left(-\frac{1}{2}\right)-6\left(\frac{2}{3}\right)+1$

$\left(x^2+1\right)\left(x\right)$

$-1-\left(2-5\right)+\left(7-4\right)$

$f\left(x\right)=x^2-5x+4$

$y'=2xy+3y-4x-6$

$0,00478\:x\:1,5$