## giúp mik bài 3b với ạ!!!! cần gấp ạ :<

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giúp mik bài 3b với ạ!!!!
cần gấp ạ :<

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1 year 2020-11-16T01:30:50+00:00 1 Answers 56 views 0

$\begin{array}{l} b)Dkxd:\left\{ \begin{array}{l} 4x + 1 \ge 0\\ 3x – 2 \ge 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x \ge – \dfrac{1}{4}\\ x \ge \dfrac{2}{3} \end{array} \right. \Rightarrow x \ge \dfrac{2}{3}\\ \sqrt {4x + 1} – \sqrt {3x – 2} = \dfrac{{x + 3}}{5}\\ \Leftrightarrow \dfrac{{4x + 1 – \left( {3x – 2} \right)}}{{\sqrt {4x + 1} + \sqrt {3x – 2} }} – \dfrac{{x + 3}}{5} = 0\\ \Leftrightarrow \dfrac{{x + 3}}{{\sqrt {4x + 1} + \sqrt {3x – 2} }} – \dfrac{{x + 3}}{5} = 0\\ \Leftrightarrow \left( {x + 3} \right)\left( {\dfrac{1}{{\sqrt {4x + 1} + \sqrt {3x – 2} }} – \dfrac{1}{5}} \right) = 0\\ \Rightarrow \left[ \begin{array}{l} x + 3 = 0\\ \sqrt {4x + 1} + \sqrt {3x – 2} = 5 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = – 3\left( {ktm} \right)\\ 4x + 1 + 2\sqrt {4x + 1} .\sqrt {3x – 2} + 3x – 2 = 25 \end{array} \right.\\ \Rightarrow 7x – 1 + 2\sqrt {12{x^2} – 5x – 2} = 25\\ \Rightarrow 2\sqrt {12{x^2} + 5x – 2} = 26 – 7x\\ \left( {dk:\dfrac{2}{3} \le x \le \dfrac{{26}}{7}} \right)\\ \Rightarrow 4\left( {12{x^2} + 5x – 2} \right) = {\left( {26 – 7x} \right)^2}\\ \Rightarrow 48{x^2} + 20x – 8 = 49{x^2} – 364x + 676\\ \Rightarrow {x^2} – 384x + 684 = 0\\ \Rightarrow \left[ \begin{array}{l} x = 1,789\left( {tm} \right)\\ x = 382,21\left( {ktm} \right) \end{array} \right.\\ Vay\,x = 1,789 \end{array}$