Bài 48:
\({x^2} + 3x\sqrt[3]{{3x + 2}} - 12 + \frac{1}{{\sqrt x }} + \frac{{\sqrt x + 8}}{x}\,\,\,\,\left( * \right)\)
Điều kiện: \(x > 0.\)
\[\begin{array}{l}\left( * \right) \Leftrightarrow {x^2} + 3x\sqrt[3]{{3x + 2}} - 12 = \frac{8}{x}\\ \Leftrightarrow {x^3} + 3{x^2}\sqrt[3]{{3x + 2}} - 12x - 8 = 0\\ \Leftrightarrow \left( {{x^3} - 8} \right) + 3x\left( {x\sqrt[3]{{3x + 2}} - 4} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + 3x.\frac{{3{x^4} + 2{x^3} - 64}}{{{x^2}\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 4x\sqrt[3]{{3x + 2}} + 16}} = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + 3x.\frac{{\left( {x - 2} \right)\left( {3{x^3} + 8{x^2} + 16x + 32} \right)}}{{{x^2}\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 4x\sqrt[3]{{3x + 2}} + 16}} = 0\\ \Leftrightarrow \left( {x - 2} \right)\left[ {{x^2} + 2x + 4 + \frac{{3x\left( {3{x^3} + 8{x^2} + 16x + 32} \right)}}{{{x^2}\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 4x\sqrt[3]{{3x + 2}} + 16}}} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 2 = 0\\{x^2} + 2x + 4 + \frac{{3x\left( {3{x^3} + 8{x^2} + 16x + 32} \right)}}{{{x^2}\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 4x\sqrt[3]{{3x + 2}} + 16}} = 0\end{array} \right.\\ \Leftrightarrow x = 2\,\,\,\left( {tm} \right)\end{array}\]
Vì \[{x^2} + 2x + 4 + \frac{{3x\left( {3{x^3} + 8{x^2} + 16x + 32} \right)}}{{{x^2}\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 4x\sqrt[3]{{3x + 2}} + 16}} > 0\,\,\forall x > 0\]
Vậy phương trình có nghiệm duy nhất \(x = 2.\)
Bài 49:
\(\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \sqrt {3x + 1} + \frac{1}{{{x^2}}} + \sqrt {x + 2} \,\,\,\left( * \right)\)
Điều kiện: \(\left\{ \begin{array}{l}x - 1 \ne 0\\x \ne 0\\3x + 1 \ge 0\\x + 2 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - \frac{1}{3}\\x \ne 1\\x \ne 0\end{array} \right.\)
\(\begin{array}{l} \Rightarrow \left( * \right) \Leftrightarrow \frac{1}{{{{\left( {x - 1} \right)}^2}}} - \frac{1}{{{x^2}}} = \sqrt {x + 2} - \sqrt {3x + 1} \\ \Leftrightarrow \frac{{{x^2} - {{\left( {x - 1} \right)}^2}}}{{{x^2}{{\left( {x - 1} \right)}^2}}} = \frac{{x + 2 - 3x - 1}}{{\sqrt {x + 2} + \sqrt {3x + 1} }}\\ \Leftrightarrow \frac{{2x - 1}}{{{x^2}{{\left( {x - 1} \right)}^2}}} = \frac{{1 - 2x}}{{\sqrt {x + 2} + \sqrt {3x + 1} }}\\ \Leftrightarrow \frac{{2x - 1}}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \frac{{2x - 1}}{{\sqrt {x + 2} + \sqrt {3x + 1} }} = 0\\ \Leftrightarrow \left( {2x - 1} \right)\left[ {\frac{1}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \frac{1}{{\sqrt {x + 2} + \sqrt {3x + 1} }}} \right] = 0\\ \Leftrightarrow 2x - 1 = 0\,\,\,\,\left( {do\,\,\,\frac{1}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \frac{1}{{\sqrt {x + 2} + \sqrt {3x + 1} }} > 0\,\,\,\forall x\,\,\,tm\,\,dkxd} \right)\\ \Leftrightarrow x = \frac{1}{2}\,\,\,\left( {tm} \right)\end{array}\)
Vậy phương trình có nghiệm duy nhất \(x = \frac{1}{2}.\)
