Question

Chứng minh đẳng thức :

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

Answer 1

Đặt VT = A = $\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}$

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

= \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\) = \(\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}\)

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

VP = B => \(B\sqrt{2}=\sqrt{2}.\sqrt{6}=2\sqrt{3}\)

=> \(A\sqrt{2}=B\sqrt{2}\Rightarrow A=B\)

Answer 2

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

<=> \(\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2\)= \(6\)

* Xét vế trái ta có :

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

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

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

=

Answer 3

Ta có \(VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{2}\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}{\sqrt{2}}=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}.\sqrt{3}=\sqrt{6}=VP\)

Vậy \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)