Question

giúp tui

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

b) \(\left(2\sqrt{3}+1\right)^2-\dfrac{1}{4}\sqrt{48}-\dfrac{2}{\sqrt{3}-1}\)

Answer 1

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

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

\(\left(2\sqrt{3}+1\right)^2-\dfrac{1}{4}\sqrt{48}-\dfrac{2}{\sqrt{3}-1}=13+4\sqrt{3}-\dfrac{1}{4}.4\sqrt{3}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=13+4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)

\(=13+3\sqrt{3}-\sqrt{3}-1=12+2\sqrt{3}\)

Answer 2

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

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

=2

b: Ta có: \(\left(2\sqrt{3}+1\right)^2-\dfrac{1}{4}\sqrt{48}-\dfrac{2}{\sqrt{3}-1}\)

\(=13-4\sqrt{3}-\sqrt{3}-1-\sqrt{3}\)

\(=12-6\sqrt{3}\)