987 B
1 (40 минут)
#семестр_1 #высшая_математика
\lim_{ x \to 3 } \frac{\sqrt{ x+13 }-2\sqrt{ x+1 }}{x^2-4x+3}
\frac{\sqrt{ x+13 }-2\sqrt{ x+1 }}{x^2-4x+3}=\frac{\sqrt{ x+13 }-2\sqrt{ x+1 }}{(x-1)(x-3)}\cdot \frac{\sqrt{ x+13 }+2\sqrt{ x+1 }}{\sqrt{ x+13 }+2\sqrt{ x+1 }}=\frac{\sqrt{ x+13 }^2-(2\sqrt{ x+1 })^2}{(x-1)(x-3)(\sqrt{ x+13 }+2\sqrt{ x+1 })}=\frac{(x+13)-4(x+1)}{(x-1)(x-3)(\sqrt{ x+13 }+2\sqrt{ x+1 })}=
=\frac{x+13-4x-4}{(x-1)(x-3)(\sqrt{ x+13 }+2\sqrt{ x+1 })}=\frac{-3x+9}{(x-1)(x-3)(\sqrt{ x+13 }+2\sqrt{ x+1 })}=\frac{-3(x-3)}{(x-1)(x-3)(\sqrt{ x+13 }+2\sqrt{ x+1 })}=\frac{-3}{(x-1)(\sqrt{ x+13 }+2\sqrt{ x+1 })}
x \equiv 3
\lim_{ x \to 3 } \frac{\sqrt{ x+13 }-2\sqrt{ x+1 }}{x^2-4x+3} = -\frac{3}{16}
6 (20 минут)
\begin{array} \\
x=1+\ln(t+2) \\
y=4t+e^{-5t}
\end{array}
x' = (\ln(t+2))'\cdot(t+2)'=\frac{1}{t+2}\cdot 1=\frac{1}{t+2}
y'=4-5e^{-5t}
\frac{y'}{x'}=4-5e^{-5t}: \frac{1}{t+2}=(4-5e^{-5t})(t+2)