rút gọn căn bậc hai

P

pinkylun

$A=\dfrac{\sqrt{2}+\sqrt{6}}{2}$

$A^2=\dfrac{2+2\sqrt{12}+6}{4}$

$A^2=\dfrac{8+4\sqrt{3}}{4}$

$A^2=2+\sqrt{3}$

$=>A=\sqrt{2+\sqrt{3}}$
 
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