Задание 25, стр 27

Question

Вычислите.
\begin{enumerate} 
\item$\frac{4+\sqrt{3}}{2+\sqrt{3}}+\frac{4-\sqrt{3}}{2-\sqrt{3}}$
\item$\frac{3+\sqrt{7}}{3-\sqrt{7}}+\frac{3-\sqrt{7}}{3+\sqrt{7}}$
\item $\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
\end{enumerate}
	extbf{Указание}: сначала избавьте от иррациональности знаменатель каждой дроби.

Gimli · Accepted Answer

$$
\begin{aligned}
1.\quad &\frac{4+\sqrt{3}}{2+\sqrt{3}}+\frac{4-\sqrt{3}}{2-\sqrt{3}}=\frac{(4+\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{(4-\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}= \frac{8-4 \sqrt{3}+2 \sqrt{3}-3+8+4 \sqrt{3}-2 \sqrt{3}-3}{4-3}=\
&= \frac{8-4 \sqrt{3}+2 \sqrt{3}-3+8+4 \sqrt{3}-2 \sqrt{3}-3}{4-3}=\frac{8-3+8-8}{4}= 8+8-6=10 .
\end{aligned}
$$
$$\begin{aligned} 2.\quad& \frac{3+\sqrt{7}}{3-\sqrt{7}}+\frac{3-\sqrt{7}}{3+\sqrt{7}}=\frac{(3+\sqrt{7})^2+(3-\sqrt{7})^2}{(3+\sqrt{7})(3-\sqrt{7})}= \frac{9+6 \sqrt{7}+7+9-6 \sqrt{7}+7}{9-7}=\frac{9+7+9+7}{2}=\frac{32}{2}=16 .\end{aligned}$$
 $$3.\quad \frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}-\frac{\sqrt{2}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{3+\sqrt{6}}{3-2}-\frac{\sqrt{6}-2}{3-2}=$$
$$=\frac{3+\sqrt{6}-(\sqrt{6}-2)}{3-2}=\frac{3+\sqrt{6}-\sqrt{6}+2}{1}=5$$

Задание 25, стр 27

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Задание 25, стр 27

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