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

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

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