Đáp án: 6,D 7.C
Giải thích các bước giải:
$\begin{array}{l}
6)\\
{9^x} - {3.3^x} + 2 = 0\\
\Rightarrow {\left( {{3^x}} \right)^2} - {2.3^x} - {3^x} + 2 = 0\\
\Rightarrow \left( {{3^x} - 2} \right)\left( {{3^x} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{3^x} = 2\\
{3^x} = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = {\log _3}2\\
x = 0
\end{array} \right.\\
\Rightarrow A = 3{\log _3}2\\
7)\\
{5.2^x} = 7.\sqrt {{{10}^x}} - {2.5^x}\\
\Rightarrow {5.2^x} - 7.\sqrt {{2^x}} .\sqrt {{5^x}} + {2.5^x} = 0\\
\Rightarrow 5.{\left( {\sqrt {{2^x}} } \right)^2} - 5.\sqrt {{2^x}} .\sqrt {{5^x}} - 2.\sqrt {{2^x}} .\sqrt {{5^x}} + 2.{\left( {\sqrt {{5^x}} } \right)^2} = 0\\
\Rightarrow 5\sqrt {{2^x}} \left( {\sqrt {{2^x}} - \sqrt {{5^x}} } \right) - 2\sqrt {{5^x}} \left( {\sqrt {{2^x}} - \sqrt {{5^x}} } \right) = 0\\
\Rightarrow \left( {\sqrt {{2^x}} - \sqrt {{5^x}} } \right)\left( {5\sqrt {{2^x}} - 2\sqrt {{5^x}} } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{2^x} = {5^x}\\
5\sqrt {{2^x}} = 2\sqrt {{5^x}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
{25.2^x} = {4.5^x}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
{\left( {\frac{2}{5}} \right)^x} = \frac{4}{{25}} \Rightarrow x = 2
\end{array} \right.\\
\Rightarrow x_1^2 + x_2^2 = 4
\end{array}$
