Tìm x: a) √ $\frac{2x-3}{x-1}$ =2 b) √4x-12 + √x-3 = $\frac{1}{3}$ × √9x-27 +4 c)2x-3 √2x-1 -5=0

0 bình luận về “Tìm x: a) √ $\frac{2x-3}{x-1}$ =2 b) √4x-12 + √x-3 = $\frac{1}{3}$ × √9x-27 +4 c)2x-3 √2x-1 -5=0”

1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   \sqrt {\dfrac{{2x – 3}}{{x – 1}}}  = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left[ \begin{array}{l}
   x \ge \dfrac{3}{2}\\
   x < 1
   \end{array} \right.} \right)\\
    \Leftrightarrow \dfrac{{2x – 3}}{{x – 1}} = 4\\
    \Leftrightarrow 2x – 3 = 4.\left( {x – 1} \right)\\
    \Leftrightarrow 2x – 3 = 4x – 4\\
    \Leftrightarrow 4x – 2x =  – 3 + 4\\
    \Leftrightarrow 2x = 1\\
    \Leftrightarrow x = \dfrac{1}{2}\,\,\,\,\left( {t/m} \right)\\
   b,\\
   \sqrt {4x – 12}  + \sqrt {x – 3}  = \dfrac{1}{3}\sqrt {9x – 27}  + 4\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 3} \right)\\
    \Leftrightarrow \sqrt {4.\left( {x – 3} \right)}  + \sqrt {x – 3}  = \dfrac{1}{3}.\sqrt {9.\left( {x – 3} \right)}  + 4\\
    \Leftrightarrow 2.\sqrt {x – 3}  + \sqrt {x – 3}  = \dfrac{1}{3}.3.\sqrt {x – 3}  + 4\\
    \Leftrightarrow 2\sqrt {x – 4}  = 4\\
    \Leftrightarrow \sqrt {x – 4}  = 2\\
    \Leftrightarrow x – 4 = 4\\
    \Leftrightarrow x = 8\\
   c,\\
   2x – 3\sqrt {2x – 1}  – 5 = 0\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{1}{2}} \right)\\
    \Leftrightarrow \left( {2x – 1} \right) – 3\sqrt {2x – 1}  – 4 = 0\\
    \Leftrightarrow \left[ {\left( {2x – 1} \right) – 4\sqrt {2x – 1} } \right] + \left[ {\sqrt {2x – 1}  – 4} \right] = 0\\
    \Leftrightarrow \sqrt {2x – 1} .\left( {\sqrt {2x – 1}  – 4} \right) + \left( {\sqrt {2x – 1}  – 4} \right) = 0\\
    \Leftrightarrow \left( {\sqrt {2x – 1}  – 4} \right)\left( {\sqrt {2x – 1}  + 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \sqrt {2x – 1}  = 4\\
   \sqrt {2x – 1}  =  – 1
   \end{array} \right.\\
    \Rightarrow \sqrt {2x – 1}  = 4\\
    \Leftrightarrow 2x – 1 = 16\\
    \Leftrightarrow x = \dfrac{{17}}{2}
   \end{array}\)

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