Đáp án đúng:
Giải chi tiết:Giải:
\(\begin{array}{l}a)\,\,\sqrt {x - 1 - 2\sqrt {x - 2} } + \sqrt {x - 2} = 1\\ \Leftrightarrow \sqrt {x - 2 - 2\sqrt {x - 2} .1 + 1} + \sqrt {x - 2} = 1\\ \Leftrightarrow \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} + \sqrt {x - 2} = 1\\ \Leftrightarrow \left| {\sqrt {x - 2} - 1} \right| = 1 - \sqrt {x - 2} .\,\,\,\,\left( * \right)\end{array}\)
ĐKXĐ: \(x-2\ge 0\Leftrightarrow x\ge 2.\)
\(\begin{array}{l}\left( * \right) \Leftrightarrow \left[ \begin{array}{l}\sqrt {x - 2} - 1 = 1 - \sqrt {x - 2} \\\sqrt {x - 2} - 1 = \sqrt {x - 2} - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2\sqrt {x - 2} = 2\\\forall \,\,x \ge 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt {x - 2} = 1\\x \ge 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x - 2 = 1\\x \ge 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\,\,\,\left( {tm} \right)\\x \ge 2\end{array} \right..\end{array}\)
Vậy phương trình đã cho có nghiệm \(x\ge 2.\)
b) \(\sqrt{x+1}+\sqrt{y+1}=\sqrt{2}\left( x+y \right).\)
ĐKXĐ: \(\left\{ \begin{array}{l}x + 1 \ge 0\\y + 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 1\\y \ge - 1\end{array} \right..\)
Đặt: \(\left\{ \begin{array}{l}a = \sqrt {x + 1} \ge 0\\b = \sqrt {y + 1} \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = {a^2} - 1\\y = {b^2} - 1\end{array} \right.,\)
khi đó giả thiết \(\Leftrightarrow a+b=\sqrt{2}\left( {{a}^{2}}+{{b}^{2}}-2 \right)\)
Ta có: \({{\left( a+b \right)}^{2}}\le 2\left( {{a}^{2}}+{{b}^{2}} \right)\)
\(\begin{array}{l} \Rightarrow 2{\left( {{a^2} + {b^2} - 2} \right)^2} \le 2\left( {{a^2} + {b^2}} \right)\\ \Leftrightarrow 2\left[ {{{\left( {{a^2} + {b^2}} \right)}^2} - 4\left( {{a^2} + {b^2}} \right) + 4} \right] - 2\left( {{a^2} + {b^2}} \right) \le 0\\ \Leftrightarrow {\left( {{a^2} + {b^2}} \right)^2} - 5\left( {{a^2} + {b^2}} \right) + 4 \le 0\\ \Leftrightarrow {\left( {{a^2} + {b^2}} \right)^2} - \left( {{a^2} + {b^2}} \right) - 4\left( {{a^2} + {b^2}} \right) + 4 \le 0\\ \Leftrightarrow \left( {{a^2} + {b^2} - 1} \right)\left( {{a^2} + {b^2} - 4} \right) \le 0\\ \Leftrightarrow 1 \le {a^2} + {b^2} \le 4.\\ \Leftrightarrow 1 - 2 \le {a^2} + {b^2} - 2 \le 4 - 2\\ \Leftrightarrow - 1 \le {a^2} + {b^2} - 2 \le 2\\ \Leftrightarrow - 1 \le P \le 2.\end{array}\)
Mà \(P=x+y=\frac{\sqrt{x+1}+\sqrt{y+1}}{\sqrt{2}}\ge 0\Rightarrow 0\le P\le 2.\)
\(\begin{array}{l} \Rightarrow Min\,\,P = 0 \Leftrightarrow \left\{ \begin{array}{l}x + y = 0\\\sqrt {x + 1} + \sqrt {y + 1} = 0\end{array} \right. \Leftrightarrow x = y = 0.\\ \Rightarrow Max\,\,P = 2 \Leftrightarrow \left\{ \begin{array}{l}x + y = 2\\\sqrt {x + 1} + \sqrt {y + 1} = 2\sqrt 2 \end{array} \right. \Leftrightarrow x = y = 1.\end{array}\)
Vậy \(Min\,\,P=0\,\,khi\,\,\,x=y=0;\,\,\,Max\,\,P=2\,\,\,khi\,\,\,x=y=1.\)
