Phương trình \({{3}^{3+3x}}+{{3}^{3-3x}}+{{3}^{4+x}}+{{3}^{4-x}}={{10}^{3}}\) có tổng các nghiệm là?
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Lời giải:
Báo sai\({{3}^{3+3x}}+{{3}^{3-3x}}+{{3}^{4+x}}+{{3}^{4-x}}={{10}^{3}}\) \(\left( 7 \right)\)
\(\left( 7 \right)\Leftrightarrow {{27.3}^{3x}}+\frac{27}{{{3}^{3x}}}+{{81.3}^{x}}+\frac{81}{{{3}^{x}}}={{10}^{3}}\Leftrightarrow 27.\left( {{3}^{3x}}+\frac{1}{{{3}^{3x}}} \right)+81.\left( {{3}^{x}}+\frac{1}{{{3}^{x}}} \right)={{10}^{3}}\text{ }\left( {{7}'} \right)\)
Đặt \(t={{3}^{x}}+\frac{1}{{{3}^{x}}}\overset{C\hat{o}si}{\mathop{\ge }}\,2\sqrt{{{3}^{x}}.\frac{1}{{{3}^{x}}}}=2\)
\(\Rightarrow {{t}^{3}}={{\left( {{3}^{x}}+\frac{1}{{{3}^{x}}} \right)}^{3}}={{3}^{3x}}+{{3.3}^{2x}}.\frac{1}{{{3}^{x}}}+{{3.3}^{x}}.\frac{1}{{{3}^{2x}}}+\frac{1}{{{3}^{3x}}}\Leftrightarrow {{3}^{3x}}+\frac{1}{{{3}^{3x}}}={{t}^{3}}-3t\)
Khi đó: \(\left( 7' \right)\Leftrightarrow 27\left( {{t}^{3}}-3t \right)+81t={{10}^{3}}\Leftrightarrow {{t}^{3}}=\frac{{{10}^{3}}}{27}\Leftrightarrow t=\frac{10}{3}>2\text{ }\left( N \right)\)
Với \(t=\frac{10}{3}\Rightarrow {{3}^{x}}+\frac{1}{{{3}^{x}}}=\frac{10}{3}\text{ }\left( {{7}''} \right)\)
Đặt \(y={{3}^{x}}>0\). Khi đó: \(\left( {7''} \right) \Leftrightarrow y + \frac{1}{y} = \frac{{10}}{3} \Leftrightarrow 3{y^2} - 10y + 3 = 0 \Leftrightarrow \left[ \begin{array}{l} y = 3{\rm{ }}\left( N \right)\\ y = \frac{1}{3}{\rm{ }}\left( N \right) \end{array} \right.\)
Với \(y = 3 \Rightarrow {3^x} = 3 \Leftrightarrow x = 1\)
Với \(y = \frac{1}{3} \Rightarrow {3^x} = \frac{1}{3} \Leftrightarrow x = - 1\)
