Question

\(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\)

Answer 1

\(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

=\(\dfrac{14}{1}=14\)

Answer 2

A = \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

= \(\dfrac{\left(7+7\right)+\left(4\sqrt{3}-4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)

vậy biểu thức A có kết quả = 14