Toán a) √(x+3) – √(x-1) < √(x-2) b) √(2+x) -1 ≥ √(1-x) 16/10/2021 By Charlie a) √(x+3) – √(x-1) < √(x-2) b) √(2+x) -1 ≥ √(1-x)
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\\sqrt {x + 3} – \sqrt {x – 1} < \sqrt {x – 2} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\ \Leftrightarrow \sqrt {x + 3} < \sqrt {x – 1} + \sqrt {x – 2} \\ \Leftrightarrow x + 3 < x – 1 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} + x – 2\\ \Leftrightarrow x + 3 < 2x – 3 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\ \Leftrightarrow 6 – x < 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\ \Leftrightarrow \left[ \begin{array}{l}x > 6\\\left\{ \begin{array}{l}x < 6\\{\left( {6 – x} \right)^2} < 4\left( {x – 1} \right)\left( {x – 2} \right)\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x > 6\\\left\{ \begin{array}{l}x < 6\\36 – 12x + {x^2} < 4{x^2} – 12x + 8\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x > 6\\\left\{ \begin{array}{l}x < 6\\3{x^2} > 28\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x < 6\\x < – \dfrac{{2\sqrt {21} }}{3}\\\dfrac{{2\sqrt {21} }}{3} < x < 6\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x < – \dfrac{{2\sqrt {21} }}{3}\\x > \dfrac{{2\sqrt {21} }}{3}\end{array} \right.\\x \ge 2 \Rightarrow S = \left( {\dfrac{{2\sqrt {21} }}{3}; + \infty } \right)\\b,\\\sqrt {2 + x} – 1 \ge \sqrt {1 – x} \,\,\,\,\,\,\,\,\,\,\left( { – 2 \le x \le 1} \right)\\ \Leftrightarrow \sqrt {2 + x} \ge \sqrt {1 – x} + 1\\ \Leftrightarrow 2 + x \ge 1 – x + 2\sqrt {1 – x} + 1\\ \Leftrightarrow 2 + x \ge 2 – x + 2\sqrt {1 – x} \\ \Leftrightarrow 2x \ge 2\sqrt {1 – x} \\ \Leftrightarrow x \ge \sqrt {1 – x} \\ \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\{x^2} \ge 1 – x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\{x^2} + x – 1 \ge 0\end{array} \right. \Leftrightarrow x \ge \dfrac{{ – 1 + \sqrt 5 }}{2}\\ \Rightarrow S = \left[ {\dfrac{{ – 1 + \sqrt 5 }}{2};1} \right]\end{array}\) Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {x + 3} – \sqrt {x – 1} < \sqrt {x – 2} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \sqrt {x + 3} < \sqrt {x – 1} + \sqrt {x – 2} \\
\Leftrightarrow x + 3 < x – 1 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} + x – 2\\
\Leftrightarrow x + 3 < 2x – 3 + 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\
\Leftrightarrow 6 – x < 2\sqrt {\left( {x – 1} \right)\left( {x – 2} \right)} \\
\Leftrightarrow \left[ \begin{array}{l}
x > 6\\
\left\{ \begin{array}{l}
x < 6\\
{\left( {6 – x} \right)^2} < 4\left( {x – 1} \right)\left( {x – 2} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 6\\
\left\{ \begin{array}{l}
x < 6\\
36 – 12x + {x^2} < 4{x^2} – 12x + 8
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 6\\
\left\{ \begin{array}{l}
x < 6\\
3{x^2} > 28
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < 6\\
x < – \dfrac{{2\sqrt {21} }}{3}\\
\dfrac{{2\sqrt {21} }}{3} < x < 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x < – \dfrac{{2\sqrt {21} }}{3}\\
x > \dfrac{{2\sqrt {21} }}{3}
\end{array} \right.\\
x \ge 2 \Rightarrow S = \left( {\dfrac{{2\sqrt {21} }}{3}; + \infty } \right)\\
b,\\
\sqrt {2 + x} – 1 \ge \sqrt {1 – x} \,\,\,\,\,\,\,\,\,\,\left( { – 2 \le x \le 1} \right)\\
\Leftrightarrow \sqrt {2 + x} \ge \sqrt {1 – x} + 1\\
\Leftrightarrow 2 + x \ge 1 – x + 2\sqrt {1 – x} + 1\\
\Leftrightarrow 2 + x \ge 2 – x + 2\sqrt {1 – x} \\
\Leftrightarrow 2x \ge 2\sqrt {1 – x} \\
\Leftrightarrow x \ge \sqrt {1 – x} \\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} \ge 1 – x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} + x – 1 \ge 0
\end{array} \right. \Leftrightarrow x \ge \dfrac{{ – 1 + \sqrt 5 }}{2}\\
\Rightarrow S = \left[ {\dfrac{{ – 1 + \sqrt 5 }}{2};1} \right]
\end{array}\)