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