Question

Chứng minh

\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4\)

Answer 1

Ta có: VT=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

=\(\sqrt{\dfrac{4-3}{\left(2+\sqrt{3}\right)^2}}+\sqrt{\dfrac{4-3}{\left(2-\sqrt{3}\right)^2}}\)

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

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

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

=\(\dfrac{4}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

=\(\dfrac{4}{4-3}\)= 4 = VP

Answer 2

Xét VT=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(=\sqrt{\dfrac{4-3}{\left(2+\sqrt{3}\right)^2}}+\sqrt{\dfrac{4-3}{\left(2-\sqrt{3}\right)^2}}\)

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

\(=\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\dfrac{4}{4-3}=4\)