### 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)
$$
\begin{array} \\
x=1+\ln(t+2) \\
y=4t+e^{-5t}
\end{array}
$$
$x' = (\ln(t+2))'\cdot(t+2)'=\frac{1}{t+2}\cdot 1=\frac{1}{t+2}$
$y'=4-5e^{-5t}$
$\frac{y'}{x'}=4-5e^{-5t}: \frac{1}{t+2}=(4-5e^{-5t})(t+2)$