Tính giá trị biểu thức $\sqrt{\frac{129}{16}+ \sqrt{2} }$

$\textit{Đáp án + Giải thích các bước giải:}$

$\sqrt{\dfrac{129}{16}+\sqrt{2}}$

$=\sqrt{8 + \dfrac{1}{16}+\sqrt{2}}$

$=\sqrt{\sqrt{8^2}+2.\sqrt{8}.\dfrac{1}{4}+\bigg(\dfrac{1}{4}\bigg)^2}$

$=\sqrt{\bigg(\sqrt{8}+\dfrac{1}{4}\bigg)^2}$

$=\bigg|\sqrt{8}+\dfrac{1}{4}\bigg|$

$=\sqrt{8}+\dfrac{1}{4}$

$=2\sqrt{2}+\dfrac{1}{4}$

   $\sqrt{\dfrac{129}{16} +\sqrt{2}}$

= $\sqrt{\dfrac{129 + 16\sqrt{2}}{16}}$

= $\dfrac{\sqrt{(8\sqrt{2})² + 2.8\sqrt{2} + 1}}{\sqrt{16}}$

= $\dfrac{\sqrt{(8√2 + 1)²}}{4}$

= $\dfrac{|8√2 + 1|}{4}$

= $\dfrac{8\sqrt{2} + 1}{4}$

= $\dfrac{8\sqrt{2}}{4}$ +  $\dfrac{1}{4}$

= 2$\sqrt{2}$ + $\dfrac{1}{4}$