Cho phương trình \({4.5^{\log (100{x^2})}} + {25.4^{\log (10x)}} = {29.10^{1 + \log x}}\).

* Hướng dẫn giải

Điều kiện \(x>0\)

\({4.5^{\log (100{x^2})}} + {25.4^{\log (10x)}} = {29.10^{1 + \log x}} \Leftrightarrow {4.25^{\log 10x}} - {29.10^{\log 10x}} + {25.4^{\log 10x}} = 0\)

\( \Leftrightarrow 4.{\left( {\frac{5}{2}} \right)^{2\log 10x}} - 29.{\left( {\frac{5}{2}} \right)^{\log 10x}} + 25 = 0 \Leftrightarrow \left[ \begin{array}{l}
{\left( {\frac{5}{2}} \right)^{\log 10x}} = 1\\
{\left( {\frac{5}{2}} \right)^{\log 10x}} = \frac{{25}}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{1}{{10}}\\
x = 10
\end{array} \right. \Rightarrow ab + 2017 = 2018\)