3^(√2(x-1)) +1) -3^x=x²-4x+3 giúp mình vs

Đáp án:

\[\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\]

Giải thích các bước giải:

ĐKXĐ:  \(x \ge 1\)

Ta có:

\(\begin{array}{l}
{3^{\sqrt {2\left( {x - 1} \right)}  + 1}} - {3^x} = {x^2} - 4x + 3\\
 \Leftrightarrow {3.3^{\sqrt {2\left( {x - 1} \right)} }} - {3.3^{x - 1}} = {x^2} - 4x + 3\\
 \Leftrightarrow {3^{\sqrt {2\left( {x - 1} \right)} }} - {3^{x - 1}} = \frac{{{x^2} - 4x + 3}}{3}\\
 \Leftrightarrow {3^{\sqrt {2\left( {x - 1} \right)} }} - {3^{x - 1}} = \frac{{{{\left( {x - 1} \right)}^2} - 2\left( {x - 1} \right)}}{3}\\
 \Leftrightarrow {3^{\sqrt {2\left( {x - 1} \right)} }} + \frac{{2\left( {x - 1} \right)}}{3} = {3^{x - 1}} + \frac{{{{\left( {x - 1} \right)}^2}}}{3}\,\,\,\,\,\,\left( 1 \right)
\end{array}\)

Xét hàm đặc trưng:  \(f\left( t \right) = {3^t} + \frac{{{t^2}}}{3}\,\,\,\,\left( {t \ge 0} \right)\) ta có:

\(f'\left( t \right) = {3^t}.\ln 3 + \frac{{2t}}{3} > 0,\,\,\,\,\forall t \ge 0\)

Do đó, \(f\left( {{t_1}} \right) = f\left( {{t_2}} \right) \Leftrightarrow {t_1} = {t_2}\,\,\,\,\,\,\left( {{t_1};{t_2} \ge 0} \right)\)

Ta có:

\(\begin{array}{l}
\left( 1 \right) \Leftrightarrow f\left( {\sqrt {2\left( {x - 1} \right)} } \right) = f\left( {x - 1} \right)\\
 \Leftrightarrow \sqrt {2\left( {x - 1} \right)}  = x - 1\\
 \Leftrightarrow 2\left( {x - 1} \right) = {\left( {x - 1} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 1 = 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\,\,\,\left( {t/m} \right)
\end{array}\)

Vậy \(\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\)