Tháng Năm 19, 2024

Cho \(A = \left( {\frac{1}{{\sqrt x + 1}} – \frac{{2\sqrt x – 2}}{{x\sqrt x – \sqrt x + x – 1}}} \right):\left( {\frac{1}{{\sqrt x – 1}} – \frac{2}{{x – 1}}} \right)\) với \(x \ge 0,x \ne 1.\) a) Rút gọn

Cho \(A = \left( {\frac{1}{{\sqrt x + 1}} – \frac{{2\sqrt x – 2}}{{x\sqrt x – \sqrt x + x – 1}}} \right):\left( {\frac{1}{{\sqrt x – 1}} – \frac{2}{{x – 1}}} \right)\) 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 = \frac{{ – 2}}{{\sqrt x + 1}}\\

b)\,\,x \in \left\{ {0;2} \right\}\\

c)\,\,\min A = – 1

\end{array}\)

B \(\begin{array}{l}

a)\,\,A = \frac{{\sqrt x – 1}}{{\sqrt x + 1}}\\

b)\,\,x = 0\\

c)\,\,\min A = – 1

\end{array}\)

C \(\begin{array}{l}

a)\,\,A = \frac{{\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 = \frac{{\sqrt x – 1}}{{\sqrt x + 1}}\\

b)\,\,x \in \left\{ {0;2} \right\}\\

c)\,\,\min A = – 1

\end{array}\)

Hướng dẫn Chọn đáp án là: D

Lời giải chi tiết:

a) Với \(x \ge 0,x \ne 1\) ta có:

\(\begin{array}{l}A = \left( {\frac{1}{{\sqrt x + 1}} – \frac{{2\sqrt x – 2}}{{x\sqrt x – \sqrt x + x – 1}}} \right):\left( {\frac{1}{{\sqrt x – 1}} – \frac{2}{{x – 1}}} \right)\\A = \left( {\frac{1}{{\sqrt x + 1}} – \frac{{2\sqrt x – 2}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\left( {\frac{1}{{\sqrt x – 1}} – \frac{2}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\A = \frac{{x – 1 – 2\sqrt x + 2}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}:\frac{{\sqrt x + 1 – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\A = \frac{{x – 2\sqrt x + 1}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}:\frac{{\sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\A = \frac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\A = \frac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} = \frac{{\sqrt x – 1}}{{\sqrt x + 1}}\end{array}\)

b) Có \(A = \frac{{\sqrt x – 1}}{{\sqrt x + 1}} = \frac{{\sqrt x + 1 – 2}}{{\sqrt x + 1}} = 1 – \frac{2}{{\sqrt x + 1}},\,\,\,\left( {x \ge 0} \right).\)

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) \(A = \frac{{\sqrt x – 1}}{{\sqrt x + 1}} = 1 – \frac{2}{{\sqrt x + 1}}\).

Ta có: \(\sqrt x + 1 \ge 1\,\,\,\left( {do\,\,\sqrt x \ge 0} \right).\) Dấu “=” xảy ra khi \(x = 0.\)

\( \Leftrightarrow \sqrt x + 1 \ge 1 \Rightarrow \frac{2}{{\sqrt x + 1}} \le \frac{2}{1} \Rightarrow – \frac{2}{{\sqrt x + 1}} \ge – 2 \Rightarrow 1 – \frac{2}{{\sqrt x + 1}} \ge – 1\)

\( \Rightarrow A \ge – 1\) dấu “=” xảy ra khi \(x = 0.\)

Vậy \(\min A = – 1\) khi \(x = 0\).