от ammornil » 23 Окт 2023, 09:56
[tex]A=15x^{2}+x-(5x^{2}-4x)-(6x^{2}+5x-2)=?, \hspace{0.5em} \because x=\frac{1}{2} \\ A=\red{15x^{2}}\blue{+x}\red{-5x^{2}}\blue{+4x}\red{-6x^{2}}\blue{-5x}+2=4x^{2}+2=4\cdot \left(\frac{1}{2}\right)^{2}+2=4\cdot \frac{1}{4}+2=1+2=3 \\ \phantom{Q} \\ B=2x^{2}y-3x-(2x^{2}y+y)+(-10x+2y)=?,\hspace{0.5em} \because x=\frac{1}{3},\hspace{0.5em} y=-1\frac{2}{3}=-\frac{5}{3} \\ B=\purple{2x^{2}y}\blue{-3x}\purple{-2x^{2}y}\orange{-y}\blue{-10x}\orange{+2y}=-13x+y=-13\cdot \frac{1}{3}+\left(-\frac{5}{3}\right)=-\frac{13}{3}-\frac{5}{3}=-\frac{18}{3}=-6[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]