Bất phương trình:\(\sqrt { - {x^2} + 6x - 5} > 8 - 2x\) có nghiệm là:
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A.
\(3 < x \le 5\).
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B.
\(2 < x \le 3\).
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C.
\( - 5 < x \le - 3\).
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D.
\( - 3 < x \le - 2\).
Sử dụng phương pháp giải bất phương trình chứa căn: \(\sqrt {f\left( x \right)} > g\left( x \right) \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}g\left( x \right) < 0\\f\left( x \right) \ge 0\end{array} \right.\\\left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) > {g^2}\left( x \right)\end{array} \right.\end{array} \right.\)
Ta có$\sqrt { - {x^2} + 6x - 5} > 8 - 2x$
$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{ - {x^2} + 6x - 5 \ge 0}\\{8 - 2x < 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{8 - 2x \ge 0}\\{ - {x^2} + 6x - 5 > {{\left( {8 - 2x} \right)}^2}}\end{array}} \right.}\end{array}} \right.$$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{1 \le x \le 5}\\{x > 4}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \le 4}\\{ - 5{x^2} + 38x - 69 > 0}\end{array}} \right.}\end{array}} \right.$$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{1 \le x \le 5}\\{x > 4}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \le 4}\\{3 < x < \dfrac{{23}}{5}}\end{array}} \right.}\end{array}} \right.$
\( \Leftrightarrow \left[ \begin{array}{l}
4 < x \le 5\\
3 < x \le 4
\end{array} \right. \Leftrightarrow 3 < x \le 5\)
Đáp án : A
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