a, x ngũ 3 – 1/4 x = 0 b, [ 2x – 1 ] tất cả ngũ 2 – [ x+ 3 ] tất cả ngũ 2 = 0 c, x ngũ 2 [ x- 3 ] + 12-4x = 0 đ,x ngũ 3 – 25x = 0 e, 3x [ x- 2 ] -4 [

0 bình luận về “a, x ngũ 3 – 1/4 x = 0 b, [ 2x – 1 ] tất cả ngũ 2 – [ x+ 3 ] tất cả ngũ 2 = 0 c, x ngũ 2 [ x- 3 ] + 12-4x = 0 đ,x ngũ 3 – 25x = 0 e, 3x [ x- 2 ] -4 [”

1. \(\begin{array}{l}
   a)\,x\left( {{x^2} – \frac{1}{4}} \right) = 0\\
   \Leftrightarrow x\left( {x – \frac{1}{2}} \right)\left( {x + \frac{1}{2}} \right) = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   x = \frac{1}{2}\\
   x = – \frac{1}{2}
   \end{array} \right.\\
   b)\,{\left( {2x – 1} \right)^2} – {\left( {x + 3} \right)^2} = 0\\
   \Leftrightarrow \left( {2x – 1 + x + 3} \right)\left( {2x – 1 – x – 3} \right) = 0\\
   \Leftrightarrow \left( {3x + 2} \right)\left( {x – 4} \right) = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   3x + 2 = 0\\
   x – 4 = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = – \frac{2}{3}\\
   x = 4
   \end{array} \right.\\
   c){x^2}\left( {x – 3} \right) – 4\left( {x – 3} \right) = 0\\
   \Leftrightarrow \left( {x – 3} \right)\left( {{x^2} – 4} \right) = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   x – 3 = 0\\
   {x^2} – 4 = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = 3\\
   {x^2} = 4
   \end{array} \right.\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = 3\\
   x = \pm 2
   \end{array} \right.\\
   d)x\left( {{x^2} – 25} \right) = 0\\
   \Leftrightarrow x\left( {x + 5} \right)\left( {x – 5} \right) = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   x = – 5\\
   x = 5
   \end{array} \right.\\
   e)\,3x\left( {x – 2} \right) + 4\left( {x – 2} \right) = 0\\
   \Leftrightarrow \left( {3x + 4} \right)\left( {x – 2} \right) = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = – \frac{4}{3}\\
   x = 2
   \end{array} \right.
   \end{array}\)

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