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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Tính B B= $\frac{\sqrt[]{2 +\sqrt[]{3}}}{2}$ : ($\frac{\sqrt[]{2 +\sqrt[]{3}}}{2}$- $\frac{2}{\sqrt[]{6}}$ +$\frac{\sqrt[]{2+\sqrt[]{

Toán Lớp 9: Tính B
B= $\frac{\sqrt[]{2 +\sqrt[]{3}}}{2}$ : ($\frac{\sqrt[]{2 +\sqrt[]{3}}}{2}$- $\frac{2}{\sqrt[]{6}}$ +$\frac{\sqrt[]{2+\sqrt[]{3}}}{2\sqrt[]{3}}$ )

Comments ( 1 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
     B = \frac{\sqrt{2 + \sqrt{3}}}{2} : (\frac{\sqrt{2 + \sqrt{3}}}{2} – \frac{2}{\sqrt{6}} + \frac{\sqrt{2 + \sqrt{3}}}{2\sqrt{3}})
        = \frac{\sqrt{2 + \sqrt{3}}}{2} : \frac{\sqrt{2 + \sqrt{3}}}{2} – \frac{\sqrt{2 + \sqrt{3}}}{2} : \frac{2}{\sqrt{6}} + \frac{\sqrt{2 + \sqrt{3}}}{2} : \frac{\sqrt{2 + \sqrt{3}}}{2\sqrt{3}}
       = \frac{\sqrt{2 + \sqrt{3}}}{2} . \frac{2}{\sqrt{2 + \sqrt{3}}} – \frac{\sqrt{2 + \sqrt{3}}}{2} . \frac{\sqrt{6}}{2} + \frac{\sqrt{2 + \sqrt{3}}}{2} . \frac{2\sqrt{3}}{\sqrt{2 + \sqrt{3}}}
       = 1 – \frac{\sqrt{12 + 6\sqrt{3}}}{4} + \sqrt{3}
       = 1 – \frac{\sqrt{9 + 2 . 3 .\sqrt{3} + 3}}{4} + \sqrt{3}
       = 1 – \frac{(3 + \sqrt{3})²}{4} + \sqrt{3}
       = 1 – \frac{3 + \sqrt{3}}{4} + \sqrt{3}
      = \frac{4 – 3 – \sqrt{3} + 4\sqrt{3}}{4}
       = \frac{1 + 3\sqrt{3}}{4}

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222-9+11+12:2*14+14 = ? ( )

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