Tìm x biết a) x ³ – 5x ² – 5x + 1 = 0 b) 3x^4 – 12(x-1) ² = 0 c) x ² + 2x – 3 = 0 d) 3x ² ( x – 3) ² – 12x ² ( x – 3) + 12x ² = 0 e) 1/4x ² ( x – 1

0 bình luận về “Tìm x biết a) x ³ – 5x ² – 5x + 1 = 0 b) 3x^4 – 12(x-1) ² = 0 c) x ² + 2x – 3 = 0 d) 3x ² ( x – 3) ² – 12x ² ( x – 3) + 12x ² = 0 e) 1/4x ² ( x – 1”

1. \[\begin{array}{l}
   a)\,\,\,{x^3} – 5{x^2} – 5x + 1 = 0\\
   \Leftrightarrow \left( {{x^3} + 1} \right) – \left( {5{x^2} + 5x} \right) = 0\\
   \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} – x + 1} \right) – 5x\left( {x + 1} \right) = 0\\
   \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} – x + 1 – 5x} \right) = 0\\
   \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} – 6x + 1} \right) = 0\\
   \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} – 6x + 9 – 8} \right) = 0\\
   \Leftrightarrow \left( {x + 1} \right)\left[ {{{\left( {x – 3} \right)}^2} – 8} \right] = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   x + 1 = 0\\
   {\left( {x – 3} \right)^2} – 8 = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = – 1\\
   {\left( {x – 3} \right)^2} = 8
   \end{array} \right.\\
   \Leftrightarrow \left[ \begin{array}{l}
   x = – 1\\
   x – 3 = 2\sqrt 2 \\
   x – 3 = – 2\sqrt 2
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = – 1\\
   x = 3 + 2\sqrt 2 \\
   x = 3 – 2\sqrt 2
   \end{array} \right..\\
   \Rightarrow S = \left\{ { – 1;\,\,3 – 2\sqrt 2 ;\,\,\,3 + 2\sqrt 2 } \right\}.\\
   b)\,\,3{x^4} – 12{\left( {x – 1} \right)^2} = 0\\
   \Leftrightarrow 3{x^4} = 12{\left( {x – 1} \right)^2}\\
   \Leftrightarrow {x^4} = 4{\left( {x – 1} \right)^2}\\
   \Leftrightarrow \left[ \begin{array}{l}
   {x^2} = 2\left( {x – 1} \right)\\
   {x^2} = – 2\left( {x – 1} \right)
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   {x^2} – 2x + 2 = 0\,\,\,\left( {VN} \right)\\
   {x^2} + 2x – 2 = 0\,\,\,\,\left( {x < 1} \right) \end{array} \right.\\ \Leftrightarrow {x^2} + 2x + 1 - 3 = 0\\ \Leftrightarrow {\left( {x + 1} \right)^2} = 3\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = \sqrt 3 \\ x + 1 = - \sqrt 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt 3 - 1\,\,\,\left( {tm} \right)\\ x = - \sqrt 3 - 1\,\,\,\left( {tm} \right) \end{array} \right.\\ \Rightarrow S = \left\{ {\sqrt 3 - 1;\,\, - \sqrt 3 - 1} \right\}.\\ c)\,\,\,{x^2} + 2x - 3 = 0\\ \Leftrightarrow {x^2} + 3x - x - 3 = 0\\ \Leftrightarrow x\left( {x + 3} \right) - \left( {x + 3} \right) = 0\\ \Leftrightarrow \left( {x + 3} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 3 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 3\\ x = 1 \end{array} \right..\\ \Rightarrow S = \left\{ { - 3;\,\,1} \right\}.\\ d)\,\,3{x^2}{\left( {x - 3} \right)^2} - 12{x^2}\left( {x - 3} \right) + 12{x^2} = 0\\ \Leftrightarrow 3{x^2}\left[ {{{\left( {x - 3} \right)}^2} - 4\left( {x - 3} \right) + 4} \right] = 0\\ \Leftrightarrow 3{x^2}{\left[ {\left( {x - 3} \right) - 2} \right]^2} = 0\\ \Leftrightarrow 3{x^2}{\left( {x - 5} \right)^2} = 0\\ \Leftrightarrow \left[ \begin{array}{l} {x^2} = 0\\ {\left( {x - 5} \right)^2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x - 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 5 \end{array} \right.\\ \Rightarrow S = \left\{ {0;\,\,5} \right\}.\\ e)\,\,\,\frac{1}{4}{x^2}{\left( {x - 1} \right)^2} - 4{\left( {x - 5} \right)^2}{x^2} = 0\\ \Leftrightarrow \frac{1}{4}{x^2}\left[ {{{\left( {x - 1} \right)}^2} - 16{{\left( {x - 5} \right)}^2}} \right] = 0\\ \Leftrightarrow {x^2}\left[ {x - 1 - 4\left( {x - 5} \right)} \right]\left[ {x - 1 + 4\left( {x - 5} \right)} \right] = 0\\ \Leftrightarrow {x^2}\left( {x - 1 - 4x + 20} \right)\left( {x - 1 + 4x - 20} \right) = 0\\ \Leftrightarrow {x^2}\left( {19 - 3x} \right)\left( {5x - 21} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 19 - 3x = 0\\ 5x - 21 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \frac{{19}}{3}\\ x = \frac{{21}}{5} \end{array} \right..\\ \Rightarrow S = \left\{ {0;\,\,\frac{{19}}{3};\,\,\frac{{21}}{5}} \right\}. \end{array}\]

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