Question

Треба розв’язок!!! допоможіть

Answer 1

Ответ:

0

Объяснение:

$\displaystyle\bf\\\sqrt{4+\sqrt{7} } =\frac{\sqrt{2} \cdot\sqrt{4+\sqrt{7} }}{\sqrt{2} } =\frac{\sqrt{8+2\sqrt{7} } }{\sqrt{2} } =\frac{\sqrt{7+2\sqrt{7}+1 } }{\sqrt{2} } =\\\\=\frac{\sqrt{(\sqrt{7}+1)^2 } }{\sqrt{2} } =\frac{|\sqrt{7}+1|}{\sqrt{2} } =\frac{\sqrt{7}+1}{\sqrt{2} }$

$\displaystyle\bf\\\sqrt{4-\sqrt{7} } =\frac{\sqrt{2} \cdot\sqrt{4-\sqrt{7} }}{\sqrt{2} } =\frac{\sqrt{8-2\sqrt{7} } }{\sqrt{2} } =\frac{\sqrt{7-2\sqrt{7}+1 } }{\sqrt{2} } =\\\\=\frac{\sqrt{(\sqrt{7}-1)^2 } }{\sqrt{2} } =\frac{|\sqrt{7}-1|}{\sqrt{2} } =\frac{\sqrt{7}-1}{\sqrt{2} }$

$\bf\\\sqrt{4+\sqrt{7} } -\sqrt{4-\sqrt{7} }-\sqrt{2} =\\\\=\dfrac{\sqrt{7} +1}{\sqrt{2} } -\dfrac{\sqrt{7} -1}{\sqrt{2} } -\sqrt{2} =\\\\=\dfrac{\sqrt{7} +1-\sqrt{7}+1 }{\sqrt{2} } -\sqrt{2} =\\\\=\dfrac{2}{\sqrt{2} } -\sqrt{2} =\sqrt{2}-\sqrt{2}=0$