4 thoughts on “phương trình

  1. \[2)\,\,\,3\sqrt {2x – 1} + x\sqrt {5 – 4{x^2}} = 4{x^2}\,\,\,(*)\]
    ĐK: \[x \in \left[ {\frac{1}{2};\frac{{\sqrt 5 }}{2}} \right]\]
    \[(*) \Leftrightarrow 3\sqrt {2x – 1} – 3\left( {2x – 1} \right) + x\sqrt {5 – 4{x^2}} + x\left( {2x – 3} \right) = 6{x^2} – 9x + 3\]
    \[ \Leftrightarrow 3\frac{{2x – 1 – {{\left( {2x – 1} \right)}^2}}}{{\sqrt {2x – 1} + 2x – 1}} + x\frac{{5 – 4{x^2} – {{\left( {2x – 3} \right)}^2}}}{{\sqrt {5 – 4{x^2}} – 2x + 3}} – \left( {6{x^2} – 9x + 3} \right) = 0\]
    \[ \Leftrightarrow \left( { – 2{x^2} + 3x – 1} \right)\left( {\frac{6}{{\sqrt {2x – 1} + 1}} + \frac{{4x}}{{\sqrt {5 – 4{x^2}} + 1}} + 3} \right) = 0\]
    \[ \Leftrightarrow – 2{x^2} + 3x – 1 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = \frac{1}{2}\end{array} \right.\]
    Vì \[\frac{6}{{\sqrt {2x – 1} + 1}} + \frac{{4x}}{{\sqrt {5 – 4{x^2}} + 1}} + 3 > 0\,\,\forall x \in \left[ {\frac{1}{2};\frac{{\sqrt 5 }}{2}} \right]\]

  2. \[3)\,\sqrt {3x – 5} + 2\sqrt[3]{{19x – 30}} = 2{x^2} – 7x + 11\,\,(*)\]
    ĐK: \[x \ge \frac{5}{3}\]
    \[(*) \Leftrightarrow \sqrt {3x – 5} – \left( {x – 1} \right) + 2\sqrt[3]{{19x – 30}} – 2x = 2{x^2} – 10x + 12\]
    \[ \Leftrightarrow \frac{{3x – 5 – {{\left( {x – 1} \right)}^2}}}{{\sqrt {3x – 5} + \left( {x – 1} \right)}} + 2\frac{{19x – 30 – {x^3}}}{{{{\left( {\sqrt[3]{{19x – 30}}} \right)}^2} + x\sqrt[3]{{19x – 30}} + {x^2}}} = 2{x^2} – 10x + 12\]
    Đến đây bạn rút nhân tử chung giống như bài 2 nhé.

  3. $$4)\,\left( {3{x^2} – 5x – 6} \right)\sqrt {2 – x} = \sqrt {3{x^2} – 6x – 5} $$
    $$ \Leftrightarrow \left( {3{x^2} – 5x – 7} \right)\sqrt {2 – x} + \left( {\sqrt {2 – x} – \sqrt {3{x^2} – 6x – 5} } \right) = 0$$
    Bạn nhân liên hợp cho phần phía sau rồi rút nhân tử chung là được

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