Question

Tính giá trị của biểu thức :
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{3}+2\right)\)

Answer 1

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

\(=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)-\sqrt{3}-2\\ =\dfrac{3\sqrt{3}+6}{3}+2\sqrt{2}-2+2-\sqrt{2}-\sqrt{3}-2\\ =\dfrac{3\left(\sqrt{3}+2\right)}{3}+\sqrt{2}-\sqrt{3}-2\\ =\sqrt{3}+2+\sqrt{2}-\sqrt{3}-2\\ =\sqrt{2}\)

Answer 2

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

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