có \(VT=\dfrac{\sqrt{1+\dfrac{2\sqrt{2}}{3}}+\sqrt{1-\dfrac{2\sqrt{2}}{3}}}{\sqrt{1+\dfrac{2\sqrt{2}}{3}}-\sqrt{1-\dfrac{2\sqrt{2}}{3}}}\)
\(=\dfrac{\sqrt{\dfrac{3+2\sqrt{2}}{3}}+\sqrt{\dfrac{3-2\sqrt{2}}{3}}}{\sqrt{\dfrac{3+2\sqrt{2}}{3}}-\sqrt{\dfrac{3-2\sqrt{2}}{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{3}}+\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{3}}}{\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{3}}-\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{3}}}\)
bạn xem lại đề ạ
\(=\dfrac{\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{3}}}{\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{3}}}\)
\(=\dfrac{\dfrac{\left|\sqrt{2}+1\right|+\left|\sqrt{2}-1\right|}{\sqrt{3}}}{\dfrac{\left|\sqrt{2}+1\right|-\left|\sqrt{2}-1\right|}{\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{2}+1-\sqrt{2}+1}\) (vì \(\sqrt{2}+1>0;\sqrt{2}-1>0\))
\(=\dfrac{2\sqrt{2}}{2}\\ =\sqrt{2}\)
