Ta có : \(\frac{3}{\sqrt{6}-\sqrt{3}}=\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{\left(\sqrt{6}-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{3}\right)}=\sqrt{6}+\sqrt{3}\)
và \(\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}=\sqrt{7}-\sqrt{3}\)
\(\Rightarrow\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\sqrt{6}+\sqrt{7}=\frac{\left(\sqrt{6}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\)
Vậy hai giá trị trên bằng nhau.
ta có +,\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{3}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}\)\(=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)
+,\(\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1\left(\sqrt{7}+\sqrt{6}\right)}{1}=\sqrt{7}+\sqrt{6}\)
vậy \(\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
