## a) √(x+3) – √(x-1) < √(x-2) b) √(2+x) -1 ≥ √(1-x)

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a) √(x+3) – √(x-1) < √(x-2) b) √(2+x) -1 ≥ √(1-x)

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1 năm 2021-10-16T01:04:04+00:00 1 Answers 5 views 0

$$\begin{array}{l} a,\\ \sqrt {x + 3} – \sqrt {x – 1} < \sqrt {x – 2} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\ \Leftrightarrow \sqrt {x + 3} < \sqrt {x – 1} + \sqrt {x – 2} \\ \Leftrightarrow x + 3 < x – 1 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} + x – 2\\ \Leftrightarrow x + 3 < 2x – 3 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\ \Leftrightarrow 6 – x < 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\ \Leftrightarrow \left[ \begin{array}{l} x > 6\\ \left\{ \begin{array}{l} x < 6\\ {\left( {6 – x} \right)^2} < 4\left( {x – 1} \right)\left( {x – 2} \right) \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x > 6\\ \left\{ \begin{array}{l} x < 6\\ 36 – 12x + {x^2} < 4{x^2} – 12x + 8 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x > 6\\ \left\{ \begin{array}{l} x < 6\\ 3{x^2} > 28 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x < 6\\ x < – \dfrac{{2\sqrt {21} }}{3}\\ \dfrac{{2\sqrt {21} }}{3} < x < 6 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x < – \dfrac{{2\sqrt {21} }}{3}\\ x > \dfrac{{2\sqrt {21} }}{3} \end{array} \right.\\ x \ge 2 \Rightarrow S = \left( {\dfrac{{2\sqrt {21} }}{3}; + \infty } \right)\\ b,\\ \sqrt {2 + x} – 1 \ge \sqrt {1 – x} \,\,\,\,\,\,\,\,\,\,\left( { – 2 \le x \le 1} \right)\\ \Leftrightarrow \sqrt {2 + x} \ge \sqrt {1 – x} + 1\\ \Leftrightarrow 2 + x \ge 1 – x + 2\sqrt {1 – x} + 1\\ \Leftrightarrow 2 + x \ge 2 – x + 2\sqrt {1 – x} \\ \Leftrightarrow 2x \ge 2\sqrt {1 – x} \\ \Leftrightarrow x \ge \sqrt {1 – x} \\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} \ge 1 – x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} + x – 1 \ge 0 \end{array} \right. \Leftrightarrow x \ge \dfrac{{ – 1 + \sqrt 5 }}{2}\\ \Rightarrow S = \left[ {\dfrac{{ – 1 + \sqrt 5 }}{2};1} \right] \end{array}$$