Question

simplify 5+root 3/7-4 root3-5+root 3/7+4 root 3

Answer 1

Dear Student,
$$\begin{gathered} \frac{5+\sqrt{3}}{7-4\sqrt{3}}-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=\frac{(5+\sqrt{3})(7+4\sqrt{3})-(7-4\sqrt{3})(5+\sqrt{3})}{\left(7-4\sqrt{3}\right){\displaystyle \left(7+4\sqrt{3}\right)}} \\ =\frac{\left(\cancel{35}+27\sqrt{3}+4\times 3\right){\displaystyle -}{\displaystyle \left(\cancel{35}-13\sqrt{3}-4\times 3\right)}}{47-16\times 3}=\frac{40\sqrt{3}+24}{-1} \\ =-(40\sqrt{3}+24) \\ =-8(3+5\sqrt{3}) \end{gathered}$$$$\begin{gathered} \frac{5+\sqrt{3}}{7-4\sqrt{3}}-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=\frac{(5+\sqrt{3})(7+4\sqrt{3})-(7-4\sqrt{3})(5+\sqrt{3})}{\left(7-4\sqrt{3}\right){\displaystyle \left(7+4\sqrt{3}\right)}} \\ =\frac{\left(\cancel{35}+27\sqrt{3}+4\times 3\right){\displaystyle -}{\displaystyle \left(\cancel{35}-13\sqrt{3}-4\times 3\right)}}{47-16\times 3}=\frac{40\sqrt{3}+24}{-1} \\ =-(40\sqrt{3}+24) \\ =-8(3+5\sqrt{3}) \end{gathered}$$