\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
Ủng hộ cách khác nè:
\(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2-\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(A^2=4-2\sqrt{4-3}\)
\(A^2=4-2=2\)
\(\Rightarrow A=\sqrt{2}\)
Vậy \(A=\sqrt{2}\)
\(=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{3}-1\right)}}{\sqrt{ }2}\)
= \(\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
= \(\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}\)
Do trời tối nên mik làm hơi tắt nha ;'')
Đặt A= √(2 + √3) - √(2 - √3)
=> √2.A= √2.√(2 + √3) - √2.√(2 - √3)
√2.A = √(4 + 2√3) - √(4 - 2√3)
√2.A = √(√3 + 1)2 - √(√3 - 1)2
√2.A= √3 + 1 - √3 +1
√2.A= 2
=> A = 2/√2=√2
