Toán ai giải hộ mình với √ {1/2- √3} + √{1/2+ √3} 25/09/2021 By Valentina ai giải hộ mình với √ {1/2- √3} + √{1/2+ √3}
Ta có $A = \sqrt{ \dfrac{1}{2-\sqrt{3}}} + \sqrt{\dfrac{1}{2 + \sqrt{3}}}$ $= \sqrt{ \dfrac{2 + \sqrt{3}}{4 – 3}} + \sqrt{ \dfrac{2 – \sqrt{3}}{4 – 3}}$ $= \sqrt{ 2 + \sqrt{3}} + \sqrt{2 – \sqrt{3}}$ Suy ra $A \sqrt{2} = \sqrt{4 + 2\sqrt{3}} + \sqrt{4 – 2\sqrt{3}}$ $= \sqrt{3 + 2.1.\sqrt{3} + 1} + \sqrt{3 – 2.1.\sqrt{3} + 1}$ $= \sqrt{(\sqrt{3} + 1)^2} + \sqrt{(\sqrt{3} – 1)^2}$ $= \sqrt{3} + 1 + \sqrt{3} – 1$ $= 2\sqrt{3}$ Suy ra $A = \sqrt{6}$. Trả lời
Ta có
$A = \sqrt{ \dfrac{1}{2-\sqrt{3}}} + \sqrt{\dfrac{1}{2 + \sqrt{3}}}$
$= \sqrt{ \dfrac{2 + \sqrt{3}}{4 – 3}} + \sqrt{ \dfrac{2 – \sqrt{3}}{4 – 3}}$
$= \sqrt{ 2 + \sqrt{3}} + \sqrt{2 – \sqrt{3}}$
Suy ra
$A \sqrt{2} = \sqrt{4 + 2\sqrt{3}} + \sqrt{4 – 2\sqrt{3}}$
$= \sqrt{3 + 2.1.\sqrt{3} + 1} + \sqrt{3 – 2.1.\sqrt{3} + 1}$
$= \sqrt{(\sqrt{3} + 1)^2} + \sqrt{(\sqrt{3} – 1)^2}$
$= \sqrt{3} + 1 + \sqrt{3} – 1$
$= 2\sqrt{3}$
Suy ra
$A = \sqrt{6}$.