Câu hỏi
Cho \(A = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{x\sqrt x - \sqrt x + x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right)\) với \(x \ge 0,x \ne 1.\)
a) Rút gọn A.
b) Tìm\(x \in Z\) để \(A \in Z\)
c) Tìm x để A đạt GTNN.
- A \(\begin{array}{l}
a)\,\,A = \dfrac{{ - 2}}{{\sqrt x + 1}}\\
b)\,\,x \in \left\{ {0;2} \right\}\\
c)\,\,\min A = - 1
\end{array}\) - B \(\begin{array}{l}
a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)\,\,x = 0\\
c)\,\,\min A = - 1
\end{array}\) - C \(\begin{array}{l}
a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)\,\,x \in \left\{ {0;\pm2} \right\}\\
c)\,\,\min A = 1
\end{array}\) - D \(\begin{array}{l}
a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)\,\,x \in \left\{ {0;2} \right\}\\
c)\,\,\min A = - 1
\end{array}\)
Lời giải chi tiết:
a) Với \(x \ge 0,x \ne 1\) ta có:
\(\begin{array}{l}A = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{x\sqrt x - \sqrt x + x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right)\\A = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\A = \dfrac{{x - 1 - 2\sqrt x + 2}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\A = \dfrac{{x - 2\sqrt x + 1}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\A = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\A = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\end{array}\)
b) Có \(A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 2}}{{\sqrt x + 1}} = 1 - \dfrac{2}{{\sqrt x + 1}},\,\,\,\left( {x \ge 0} \right).\)
Đặt \(B = \sqrt x + 1\), để A nguyên khi x nguyên thì B là ước nguyên của 2.
Có \(B > 0\,\,do\,x \ge 0 \Rightarrow B = \left\{ {1;2} \right\}.\)
TH1: \(\sqrt x + 1 = 1 \Leftrightarrow x = 0\,\,\left( {tm} \right)\)
TH2: \(\sqrt x + 1 = 2 \Leftrightarrow x = 1\,\,\left( {tm} \right)\)
Vậy \(x = \left\{ {0;2} \right\}\) thì A nguyên.
c) \(A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = 1 - \dfrac{2}{{\sqrt x + 1}}\).
Ta có: \(\sqrt x + 1 \ge 1\,\,\,\left( {do\,\,\sqrt x \ge 0} \right).\) Dấu “=” xảy ra khi \(x = 0.\)
\( \Leftrightarrow \sqrt x + 1 \ge 1 \Rightarrow \dfrac{2}{{\sqrt x + 1}} \le \dfrac{2}{1} \Rightarrow - \dfrac{2}{{\sqrt x + 1}} \ge - 2 \Rightarrow 1 - \dfrac{2}{{\sqrt x + 1}} \ge - 1\)
\( \Rightarrow A \ge - 1\) dấu “=” xảy ra khi \(x = 0.\)
Vậy \(\min A = - 1\) khi \(x = 0\).
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