cho A= $\frac{x\sqrt[]{x}-4x-\sqrt[]{x}+4}{2x\sqrt[]{x}-14x+28\sqrt[]{x}-16}$ tìm đkxđ rút gọn A tìm x nguyên để A nguyên cho Q= $\frac{2\sqrt[]{x}-

0 bình luận về “cho A= $\frac{x\sqrt[]{x}-4x-\sqrt[]{x}+4}{2x\sqrt[]{x}-14x+28\sqrt[]{x}-16}$ tìm đkxđ rút gọn A tìm x nguyên để A nguyên cho Q= $\frac{2\sqrt[]{x}-”

1. Giải thích các bước giải:
   \(\begin{array}{l}
   1.A = \frac{{x\sqrt x – 4x – \sqrt x + 4}}{{2x\sqrt x – 14x + 28\sqrt x – 16}}\\
   Dkxd\left\{ \begin{array}{l}
   x \ge 0\\
   2x\sqrt x – 14x + 28\sqrt x – 16 \ne 0
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   x \ge 0\\
   x \ne 1\\
   x \ne 4\\
   x \ne 16
   \end{array} \right.\\
   A = \frac{{x(\sqrt x – 4) – (\sqrt x – 4)}}{{(\sqrt x – 1)(\sqrt x – 2)(\sqrt x – 4)}} = \frac{{(\sqrt x – 1)(\sqrt x + 1)(\sqrt x – 4)}}{{(\sqrt x – 1)(\sqrt x – 2)(\sqrt x – 4)}} = \frac{{\sqrt x + 1}}{{\sqrt x – 2}}\\
   A = \frac{{\sqrt x + 1}}{{\sqrt x – 2}} = 1 + \frac{3}{{\sqrt x – 2}}\\
   A\;nguyen \Leftrightarrow 3 \vdots (\sqrt x – 2) \Leftrightarrow (\sqrt x – 2) \in \left\{ { \pm 1; \pm 3} \right\} \Rightarrow x \in \left\{ {1;9;25} \right\}\\
   2.Q = \frac{{2\sqrt x – 9}}{{x – 5\sqrt x + 6}} – \frac{{\sqrt x + 3}}{{\sqrt x – 2}} – \frac{{2\sqrt x + 1}}{{3 – \sqrt x }}\\
   Dk\left\{ {\begin{array}{*{20}{l}}
   {x \ge 0}\\
   {x – 5\sqrt x + 6 \ne 0}\\
   {\sqrt x – 2 \ne 0}\\
   {3 – \sqrt x \ne 0}
   \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
   {x \ge 0}\\
   {x \ne 4}\\
   {x \ne 9}
   \end{array}} \right.\\
   Q = \frac{{2\sqrt x – 9}}{{(\sqrt x – 2)(\sqrt x – 3)}} – \frac{{(\sqrt x + 3)(\sqrt x – 3)}}{{(\sqrt x – 2)(\sqrt x – 3)}} + \frac{{(2\sqrt x + 1)(\sqrt x – 2)}}{{(\sqrt x – 2)(\sqrt x – 3)}} = \frac{{2\sqrt x – 9 – x + 9 + 2x – 3\sqrt x – 2}}{{(\sqrt x – 2)(\sqrt x – 3)}} = \frac{{x – \sqrt x – 2}}{{(\sqrt x – 2)(\sqrt x – 3)}} = \frac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
   Q < 1 \Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1 \Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - 1 < 0 \Leftrightarrow \frac{4}{{\sqrt x - 3}} < 0 \Rightarrow \sqrt x - 3 < 0 \Leftrightarrow x < 9\\ Ket\;hop\;dk \Rightarrow \left\{ \begin{array}{l} 0 \le x < 9\\ x \ne 4 \end{array} \right.\\ 3.P = \left( {1 - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right):\left( {\frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}} + \frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }}} \right)\\ Dk:\left\{ \begin{array}{l} x \ge 0\\ \sqrt x - 2 \ne 0\\ 3 - \sqrt x \ne 0\\ x - 5\sqrt x + 6 \ne 0\\ \sqrt x + 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {x \ge 0}\\ {x \ne 4}\\ {x \ne 9} \end{array}} \right.\\ P = \frac{1}{{\sqrt x + 1}}:\left( {\frac{{\sqrt x + 2}}{{(\sqrt x - 2)(\sqrt x - 3)}} + \frac{{(\sqrt x + 3)(\sqrt x - 3)}}{{(\sqrt x - 2)(\sqrt x - 3)}} - \frac{{(\sqrt x + 2)(\sqrt x - 2)}}{{(\sqrt x - 2)(\sqrt x - 3)}}} \right) = \frac{1}{{\sqrt x + 1}}:\frac{{\sqrt x + 2 + x - 9 - x + 4}}{{(\sqrt x - 2)(\sqrt x - 3)}} = \frac{1}{{\sqrt x + 1}}:\frac{{\sqrt x - 3}}{{(\sqrt x - 2)(\sqrt x - 3)}} = \frac{{\sqrt x - 2}}{{\sqrt x + 1}}\\ Khi\;x = 4 - 2\sqrt 3 \Rightarrow \sqrt x = \sqrt {4 - 2\sqrt 3 } = \sqrt {3 - 2\sqrt 3 + 1} = \sqrt {{{(\sqrt 3 - 1)}^2}} = \sqrt 3 - 1\\ \Rightarrow P = \frac{{\sqrt 3 - 1 - 2}}{{\sqrt 3 - 1 + 1}} = 1 - \sqrt 3 \end{array}\)

   Trả lời