Question

[Bùi Tuấn Trung]

Chứng minh rằng:\(\left(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\right)\): \(\dfrac{1}{\sqrt{2}}=2\sqrt{2}\) 

Answer 1

[HT.Phong (9A5)]

Ta có:

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

\(=\left[\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]:\dfrac{1}{\sqrt{2}}\)

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

\(=\dfrac{4}{2}:\dfrac{1}{\sqrt{2}}\)

\(=2:\dfrac{1}{\sqrt{2}}\)

\(=2\sqrt{2}\left(dpcm\right)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

\(VT=\left(\sqrt{3}+1-\sqrt{3}+1\right)\cdot\sqrt{2}=2\cdot\sqrt{2}=VP\)