Giải phương trình: \[8{x^2} - 13x + 11

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Giải phương trình: \[8{x^2} - 13x + 11 = \frac{2}{x} + \left( {1 + \frac{3}{x}} \right)\sqrt[3]{{3{x^2} - 2}}.\]

Nhập câu trả lời ở đây

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Giải thích

\[8{x^2} - 13x + 11 = \frac{2}{x} + \left( {1 + \frac{3}{x}} \right)\sqrt[3]{{3{x^2} - 2}}.\](ĐKXĐ: \[x \ne 0\])

PT đã cho \[ \Leftrightarrow x\left( {8{x^2} - 13x + 11} \right) = 2 + \left( {x + 3} \right)\sqrt[3]{{3{x^2} - 2}}\]

\[ \Leftrightarrow 8{x^3} - 13{x^2} + 11x = 2 + \left( {x + 3} \right)\sqrt[3]{{3{x^2} - 2}}\]

\[ \Leftrightarrow \left( {\left. {8{x^3} - 15{x^2} + 6x + 1} \right)} \right. + \left( {x + 3} \right)\left( {2x - 1 - \sqrt[3]{{3{x^2} - 2}}} \right) = 0\]

\[ \Leftrightarrow {\left( {x - 1} \right)^2}\left( {8x + 1} \right) + \frac{{\left( {x + 3} \right)\left[ {{{\left( {2x - 1} \right)}^3} - \left( {3{x^2} - 2} \right)} \right]}}{{{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}} = 0\]

\[ \Leftrightarrow {\left( {x - 1} \right)^2}\left( {8x + 1} \right)\left[ {1 + \frac{{\left( {x + 3} \right)}}{{{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}}} \right] = 0\]

\[ \Leftrightarrow {\left( {x - 1} \right)^2}\left( {8x + 1} \right)\left[ {\frac{{\left( {x + 3} \right) + {{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}}{{{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}}} \right] = 0\]

\[ \Leftrightarrow {\left( {x - 1} \right)^2}\left( {8x + 1} \right)\frac{{{{\left[ {\sqrt[3]{{3{x^2} - 2}} + \frac{1}{2}\left( {2x - 1} \right)} \right]}^2} + \frac{1}{4}\left[ {12{{\left( {x - \frac{1}{3}} \right)}^2} + \frac{{41}}{3}} \right]}}{{{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}} = 0\]

Dễ thấy \[\frac{{{{\left[ {\sqrt[3]{{3{x^2} - 2}} + \frac{1}{2}\left( {2x - 1} \right)} \right]}^2} + \frac{1}{4}\left[ {12{{\left( {x - \frac{1}{3}} \right)}^2} + \frac{{41}}{3}} \right]}}{{{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)\sqrt[3]{{3{x^2} - 2}} + \sqrt[3]{{{{\left( {3{x^2} - 2} \right)}^2}}}}} \ge 0,\forall x \ne 0\]

Từ đó suy ra \[{\left( {x - 1} \right)^2}\left( {8x + 1} \right) = 0 \Rightarrow x \in \left\{ {1; - \frac{1}{8}} \right\}\] .vậy tập nghiệm S=\[\left\{ {1; - \frac{1}{8}} \right\}\]