Giải hệ phương trình: \left\{ \begin{array}{l} x - 4\sqrt {x - 1} + y - \frac{{2\left( {{y^2} + 24} \right)}}{{2{y^2} - 1}} = 0\left( 1 \right)\\ \sqrt {5x + y - 5} + \sqrt {1 - x + y} = 6\left( 2 \right) \end{array} \right.
Giải:
\begin{array}{l} \left( 2 \right) \Leftrightarrow 5 - \sqrt {5x + y - 5} + 1 - \sqrt {1 - x + y} = 0 \Leftrightarrow \dfrac{{30 - 5x - y}}{{5 + \sqrt {5x + y - 5} }} + \dfrac{{x - y}}{{1 + \sqrt {1 - x + y} }} = 0\\ \Leftrightarrow \dfrac{{5x + y - 30}}{{5 + \sqrt {5x + y - 5} }} = \dfrac{{x - y}}{{1 + \sqrt {1 - x + y} }} = \dfrac{{5x + y - 30 + x - y}}{{5 + \sqrt {5x + y - 5} + 1 + \sqrt {1 - x + y} }} = \dfrac{{x - 5}}{2}\,\,\,\,\,\,\\ \Leftrightarrow 2\left( {x - y} \right) = \left( {x - 5} \right) + \left( {x - 5} \right)\sqrt {1 - x + y} .\,\,Coi\,\left\{ \begin{array}{l} a = \sqrt {1 - \left( {x - y} \right)} \left( {a \ge 0} \right) \Leftrightarrow x - y = 1 - {a^2}\\ b = x - 5 \end{array} \right.\\ \Rightarrow 2\left( {1 - {a^2}} \right) = b + ab \Leftrightarrow 2\left( {a - 1} \right)\left( {a + 1} \right) + b\left( {a + 1} \right) = 0\, \Leftrightarrow \left( {a + 1} \right)\left[ {2\left( {a - 1} \right) + b} \right] = 0\\ \Leftrightarrow 2\left( {a - 1} \right) + b = 0 \Leftrightarrow 2\sqrt {1 - x + y} = 7 - x \Leftrightarrow 4 - 4x + 4y = {x^2} - 14x + 49\\ \Leftrightarrow 4y = {x^2} - 10x + 45 = {\left( {x - 5} \right)^2} + 20 \ge 20 \Leftrightarrow y \ge 5\\ \left( 1 \right) \Leftrightarrow \left( {x - 1 - 4\sqrt {x - 1} + 4} \right) + y - 3 - \dfrac{{2\left( {{y^2} + 24} \right)}}{{2{y^2} - 1}} = 0 \Leftrightarrow {\left( {\sqrt {x - 1} - 2} \right)^2} + \dfrac{{\left( {y - 5} \right)\left( {2{y^2} + 2y + 9} \right)}}{{2{y^2} - 1}} = 0\left( * \right)\,\\ Do\,\left\{ \begin{array}{l} {\left( {\sqrt {x - 1} - 2} \right)^2} \ge 0\\ \dfrac{{\left( {y - 5} \right)\left( {2{y^2} + 2y + 9} \right)}}{{2{y^2} - 1}} \ge 0\left( {Do\,y \ge 5} \right) \end{array} \right. \Rightarrow V{T_{\left( * \right)}} \ge 0 \Rightarrow \left( * \right) \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 1} - 2 = 0\\ y = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 5\\ y = 5 \end{array} \right. \end{array}
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