## Giải cách phương trình sau

Question

Giải cách phương trình sau

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1 year 2020-11-27T04:55:38+00:00 1 Answers 47 views 0

$\begin{array}{l} 1)\sqrt {16{x^2}} = 7\\ \Leftrightarrow 16{x^2} = 49\\ \Leftrightarrow {x^2} = \dfrac{{49}}{{16}}\\ \Leftrightarrow x = \pm \dfrac{7}{4}\\ 2)\sqrt {25{{\left( {x – 1} \right)}^2}} = 50\\ \Leftrightarrow 25{\left( {x – 1} \right)^2} = 2500\\ \Leftrightarrow {\left( {x – 1} \right)^2} = 100\\ \Leftrightarrow \left[ \begin{array}{l} x – 1 = 10\\ x – 1 = – 10 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 11\\ x = – 9 \end{array} \right.\\ 3)\sqrt {4\left( {{x^2} – 4x + 4} \right)} = 12\\ \Leftrightarrow \sqrt {{2^2}{{\left( {x – 2} \right)}^2}} = 12\\ \Leftrightarrow \left[ \begin{array}{l} 2\left( {x – 2} \right) = 12\\ 2\left( {x – 2} \right) = – 12 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x – 2 = 6\\ x – 2 = – 6 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 8\\ x = – 4 \end{array} \right.\\ 4)\sqrt {4{x^2} + 4x + 1} – 6 = 0\\ \Leftrightarrow \sqrt {{{\left( {2x + 1} \right)}^2}} = 6\\ \Leftrightarrow \left| {2x + 1} \right| = 6\\ \Leftrightarrow \left[ \begin{array}{l} 2x + 1 = 6\\ 2x + 1 = – 6 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{5}{2}\\ x = – \dfrac{7}{2} \end{array} \right.\\ 5)\sqrt {5\left( {2x – 1} \right)} = 15\\ \Leftrightarrow 5\left( {2x – 1} \right) = 225\\ \Leftrightarrow 2x – 1 = 45\\ \Leftrightarrow x = 23\\ 6)\sqrt {3x – 5} + 4 = 7\\ \Leftrightarrow \sqrt {3x – 5} = 3\\ \Leftrightarrow 3x – 5 = 9\\ \Leftrightarrow x = \dfrac{{14}}{3}\\ 7)\sqrt {1 – 4x} – 3 = 7\\ \Leftrightarrow \sqrt {1 – 4x} = 10\\ \Leftrightarrow 1 – 4x = 100\\ \Leftrightarrow x = \dfrac{{ – 99}}{4}\\ 8)\sqrt {{x^2} – 6x + 9} – 9 = 0\\ \Leftrightarrow \sqrt {{{\left( {x – 3} \right)}^2}} = 9\\ \Leftrightarrow \left| {x – 3} \right| = 9\\ \Leftrightarrow \left[ \begin{array}{l} x – 3 = 9\\ x – 3 = – 9 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 12\\ x = – 6 \end{array} \right.\\ 9)\sqrt { – 5x + 3} = 2\\ \Leftrightarrow – 5x + 3 = 4\\ \Leftrightarrow x = – \dfrac{1}{5}\\ 10)7 – \sqrt {x – 1} = 10\\ \Leftrightarrow \sqrt {x – 1} = – 3\left( {vo\,nghiem} \right) \end{array}$