Question

кто знает как решить?​

Answer 1

$y=\frac{2x-3x^{2}}{2-x}\\\\f(3)=\frac{2*3-3*3^{2}}{2-3}=\frac{6-27}{-1}=\frac{-21}{-1}=21\\\\f(-2)=\frac{2*(-2)-3*(-2)^{2}}{2-(-2)}=\frac{-4-12}{2+2}=\frac{-16}{4}=-4$

Answer 2

$y=\frac{2x-3x^2}{2-x}\\\\f(3)=\frac{2\cdot 3-3\cdot 3^2}{2-3}=\frac{6-27}{-1}=21\\\\f(-2)=\frac{2\cdot (-2)-3\cdot (-2)^2}{2-(-2)}=\frac{-4-6}{4}=\frac{-16}{4}=-4$