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

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    2021-10-16T01:05:25+00:00

    Giải thích các bước giải:

     Ta có:

    \(\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}\)

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