\( b)2000 - M = 2000 - \dfrac{1}{{1 - \sqrt x + x}}\\ x - \sqrt x + 1 = x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \)
Với \(x \ge 4\) \(\Rightarrow \sqrt x \ge 2\)
\( \Rightarrow \sqrt x - \dfrac{1}{2} \ge \dfrac{3}{2}\\ \Rightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge \dfrac{9}{4}\\ \Rightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{9}{4} + \dfrac{3}{4} = 3\\ \Rightarrow x - \sqrt x + 1 \ge 3\\ \Rightarrow \dfrac{1}{{1 - \sqrt x + x}} \le \dfrac{1}{3}\\ \Rightarrow 2000 - \dfrac{1}{{1 - \sqrt x + x}} \ge 2000 - \dfrac{1}{3} = \dfrac{{5999}}{3} \)
Vậy với \(x \ge 4\) thì $Min$ \(2000 - M = \dfrac{{5999}}{3}\) khi và chỉ khi \(x=4\)
$c)$ \(M = \dfrac{1}{{1 - \sqrt x + x}}\)
Để $M$ nguyên \(\Rightarrow1-\sqrt{x}+x\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}1-\sqrt{x}+1=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\\1-\sqrt{x}+1=-1\Leftrightarrow x-\sqrt{x}+2=0\left(VN\right)\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) thì $M$ nguyên
\(a)\) $M$ có nghĩa \(\left\{ {\begin{array}{*{20}{c}} {x \ge 0}\\ {1 - x \ne 0}\\ {2\sqrt x - 1 \ne 0} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x \ge 0}\\ {x \ne 1}\\ {x \ne \dfrac{1}{4}} \end{array}} \right.\)
\( M = 1 - \left[ {\dfrac{{2x - 1 + \sqrt x }}{{1 - x}} + \dfrac{{2x\sqrt x + x - \sqrt x }}{{1 + x\sqrt x }}} \right]\left[ {\dfrac{{\left( {x - \sqrt x } \right)\left( {1 - \sqrt x } \right)}}{{2\sqrt x - 1}}} \right]\\ M = 1 - \left[ {\dfrac{{2x + 2\sqrt x - \sqrt x - 1}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{\sqrt x \left( {2x + 2\sqrt x - \sqrt x - 1} \right)}}{{\left( {1 + \sqrt x } \right)\left( {1 - \sqrt x + x} \right)}}} \right].\dfrac{{\left( {x - \sqrt x } \right)\left( {1 - \sqrt x } \right)}}{{2\sqrt x - 1}}\\ M = 1 - \left[ {\dfrac{{2\sqrt x - 1}}{{1 - \sqrt x }} + \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)}}{{1 - \sqrt x + x}}} \right].\dfrac{{\left( {x - \sqrt x } \right)\left( {1 - \sqrt x } \right)}}{{2\sqrt x - 1}}\\ M = 1 - \dfrac{{\left( {1 - \sqrt x + x} \right)\left( {2\sqrt x - 1} \right) + \left( {1 - \sqrt x } \right)\left( {2\sqrt x - 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 - \sqrt x + x} \right)}}.\dfrac{{\left( {x - \sqrt x } \right)\left( {1 - \sqrt x } \right)}}{{2\sqrt x - 1}}\\ M = 1 - \dfrac{{2\sqrt x - 1}}{{1 - \sqrt x + x}}.\dfrac{{x - \sqrt x }}{{2\sqrt x - 1}}\\ M = 1 - \dfrac{1}{{1 - \sqrt x + x}}.\left( {x - \sqrt x } \right)\\ M = 1 - \dfrac{{x - \sqrt x }}{{1 - \sqrt x + x}}\\ M = \dfrac{{1 - \sqrt x + x - \left( {x - \sqrt x } \right)}}{{1 - \sqrt x + x}}\\ M = \dfrac{1}{{1 - \sqrt x + x}} \)
Gửi câu $a)$ trước nhé! Muốn tẩu hỏa rồi... đợi chút
