Đáp án:
$D.\, \dfrac{39\sqrt7}{5}$
Giải thích các bước giải:
$\begin{array}{l} \dfrac{1}{5^{\displaystyle{-x-1}}} + 3{\sqrt5}^{\displaystyle{2x}}-25^{\displaystyle{\dfrac{x-1}{2}}}\\ =5^{\displaystyle{x+1}}+3.\left(5^{\dfrac12}\right)^{\displaystyle{2x}} - (5^2)^{\displaystyle{\dfrac{x-1}{2}}}\\ = 5.5^{\displaystyle{x}}+ 3.5^{\displaystyle{x}} -5^{\displaystyle{x-1}}\\ =8.5^{\displaystyle{x}}-\dfrac{1}{5}\cdot5^{\displaystyle{x}}\\ = \dfrac{39}{5}\cdot5^{\displaystyle{x}}\\ Với\,\,5^x=\sqrt7\,\,ta\,\,được:\\ \quad \dfrac{39}{5}\cdot\sqrt7 = \dfrac{39\sqrt7}{5} \end{array}$
