Question

\(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{4-2\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

Answer 1

Ta có : \(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{4-2\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

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

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

Answer 2

\(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{4-2\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

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

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

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