Question

thực hiện phép tính

a, \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)

Answer 1

\(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\\ =\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\\ =\dfrac{6}{7}\)

Answer 2

=\(\dfrac{3-\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}+\dfrac{3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\)

=\(\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\)

=\(\dfrac{3+3}{9-2}\)

=\(\dfrac{6}{7}\)