化簡:\( \frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}} \)
已知
\( \frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}} \)
要求:
我們必須化簡給定的表示式。
解答
我們知道:
分母為${\sqrt{a}}$的分式的有理化因子是${\sqrt{a}}$。
分母為${\sqrt{a}-\sqrt{b}}$的分式的有理化因子是${\sqrt{a}+\sqrt{b}}$。
分母為${\sqrt{a}+\sqrt{b}}$的分式的有理化因子是${\sqrt{a}-\sqrt{b}}$。
因此:
$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$
$=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$
$=\frac{2(\sqrt{5}-\sqrt{3})}{5-3}$
$=\frac{2(\sqrt{5}-\sqrt{3})}{2}$
$=\sqrt{5}-\sqrt{3}$
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1 \times(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
$=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}$
$=\frac{\sqrt{3}-\sqrt{2}}{1}$
$=\sqrt{3}-\sqrt{2}$
$\frac{3}{\sqrt{5}+\sqrt{2}}=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$
$=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$
$=\frac{3(\sqrt{5}-\sqrt{2})}{5-2}$
$=\frac{3(\sqrt{5}-\sqrt{2})}{3}$
$=\sqrt{5}-\sqrt{2}$
因此:
$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}=(\sqrt{5}-\sqrt{3})+(\sqrt{3}-\sqrt{2})-(\sqrt{5}-\sqrt{2})$
$=\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}$
$=0$
因此,$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}=0$。