Hãy rút gọn: \(A = {1 \over {1 - 5x}}.\sqrt {3{x^2}\left( {25{x^2} - 10x + 1} \right)} ;\)\(\,\,\,\,\,\,0 \le x...

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}
A = \frac{1}{{1 - 5x}}.\sqrt {3{x^2}\left( {25{x^2} - 10x + 1} \right)} \\
= \frac{1}{{1 - 5x}}.\sqrt 3 .\sqrt {{x^2}{{\left( {5x - 1} \right)}^2}}
\end{array}\)

\(\eqalign{  & = {{\sqrt 3 } \over {1 - 5x}}\left| x \right|.\left| {5x - 1} \right|  \cr  & \text{Vì }\,x \ge 0 \Rightarrow \left| x \right| = x  \cr  & \text{Vì }\,x < {1 \over 5} \Rightarrow 5x - 1 < 0 \cr&\Rightarrow \left| {5x - 1} \right| = 1 - 5x  \cr  & Vậy\,:\,\,A = x\sqrt 3  \cr} \)