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Toán Học · Lớp $9$
$...$ $B=\sqrt{\frac{11\sqrt5-3\sqrt{40}}{\sqrt5}}-\sqrt{4-\sqrt7}-\frac{11}{\sqrt{14}+6}$

Question

Huycindy · Accepted Answer

$B = \sqrt{\dfrac{11\sqrt{5}-3\sqrt{40}}{\sqrt{5}}} - \sqrt{4-\sqrt{7}} - \dfrac{11}{\sqrt{14}+6}$
$B = \sqrt{11 - \dfrac{3\sqrt{40}}{\sqrt{5}}} - \sqrt{4 - \sqrt{7}} - \dfrac{11}{\sqrt{14} + 6}$
$B = \sqrt{11 - 3\sqrt{8}} - \sqrt{4 - \sqrt{7}} - \dfrac{11 \cdot \left(\sqrt{14} - 6\right)}{14 - 36}$
$B = \sqrt{11 - 6\sqrt{2}} - \sqrt{4 - \sqrt{7}} - \dfrac{11 \cdot \left(\sqrt{14} - 6\right)}{-22}$
$B = \sqrt{\left(3 - \sqrt{2}\right)^2} - \sqrt{4 - \sqrt{7}} + \dfrac{\sqrt{14} - 6}{2}$
$B = 3 - \sqrt{2} - \dfrac{\sqrt{8 - 2\sqrt{7}}}{\sqrt{2}} + \dfrac{\sqrt{14} - 6}{2}$
$B = 3 - \sqrt{2} - \dfrac{\sqrt{\left(\sqrt{7} - 1\right)^2}}{\sqrt{2}} + \dfrac{\sqrt{14} - 6}{2}$
$B = 3 - \sqrt{2} - \dfrac{\sqrt{7} - 1}{\sqrt{2}} + \dfrac{\sqrt{14} - 6}{2}$
$B = 3 - \sqrt{2} - \dfrac{\sqrt{14} - \sqrt{2}}{2} + \dfrac{\sqrt{14} - 6}{2}$
$B = \dfrac{6 - 2\sqrt{2} - \sqrt{14} + \sqrt{2} + \sqrt{14} - 6}{2}$
$B = -\dfrac{\sqrt{2}}{2}$

Timi · Answer

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Hurricane · Answer

{"userId":5433079,"userName":"Minh Thư ୨୧"}

$B=\sqrt{\dfrac{11\sqrt{5}-3\sqrt{40}}{\sqrt{5}}}-\sqrt{4-\sqrt{7}}-\dfrac{11}{\sqrt{14}+6}$

$B=\sqrt{\dfrac{11\sqrt{5}-6\sqrt{10}}{\sqrt{5}}}-\sqrt{4-\sqrt{7}}-\dfrac{11}{\sqrt{14}+6}$

$B=\sqrt{11-6\sqrt{2}}-\sqrt{4-\sqrt{7}}-\dfrac{11}{\sqrt{14}+6}$

$B=3-\sqrt{2}-\sqrt{4-\sqrt{7}}-\dfrac{11}{\sqrt{14}+6}$

$B=3-\sqrt{2}-\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)-\dfrac{11}{\sqrt{14}+6}$

$B=3-\sqrt{2}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\dfrac{11}{\sqrt{14}+6}$

$B=3-\sqrt{2}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\dfrac{6-\sqrt{14}}{2}$

$B=3-\sqrt{2}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-3+\dfrac{\sqrt{14}}{2}$

$B=-\sqrt{2}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\dfrac{\sqrt{14}}{2}$

$B=-\sqrt{2}+\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{14}}{2}+\dfrac{\sqrt{14}}{2}$

$B=-\sqrt{2}+\dfrac{\sqrt{2}}{2}$

$B=-\dfrac{\sqrt{2}}{2}$

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