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Giải các bất phương trình:
LG a
\(\sqrt {{x^2} + x - 6} < x - 1\)
Phương pháp giải:
Biến đổi tương đương
\(\sqrt f < g \Leftrightarrow \left\{ \begin{array}{l}
g > 0\\
0 \le f < {g^2}
\end{array} \right.\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sqrt {{x^2} + x - 6} < x - 1\cr& \Leftrightarrow \left\{ \matrix{
{x^2} + x - 6 \ge 0 \hfill \cr
x - 1 > 0 \hfill \cr
{x^2} + x - 6 < {(x - 1)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \le - 3\\x \ge 2\end{array} \right.\\x > 1\\{x^2} + x - 6 < {x^2} - 2x + 1\end{array} \right.\cr &\Leftrightarrow \left\{ \matrix{\left[ \matrix{x \le -3 \hfill \cr x \ge 2 \hfill \cr} \right. \hfill \cr x > 1 \hfill \cr 3x < 7 \hfill \cr} \right. \Leftrightarrow 2 \le x < {7 \over 3} \cr} \)
Vậy \(S = {\rm{[}}2,{7 \over 3})\)
LG b
\(\sqrt {2x - 1} \le 2x - 3\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sqrt {2x - 1} \le 2x - 3 \cr &\Leftrightarrow \left\{ \matrix{
2x - 1 \ge 0 \hfill \cr
2x - 3 \ge 0 \hfill \cr
2x - 1 \le {(2x - 3)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \begin{array}{l}2x \ge 1\\2x \ge 3\\2x - 1 \le 4{x^2} - 12x + 9\end{array} \right.\cr &\Leftrightarrow \left\{ \matrix{x \ge {1 \over 2} \hfill \cr x \ge {3 \over 2} \hfill \cr 4{x^2} - 14x + 10 \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{x \ge {3 \over 2} \hfill \cr \left[ \matrix{x \le 1 \hfill \cr x \ge {5 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {5 \over 2} \cr} \)
Vậy \(S = {\rm{[}}{5 \over 2}; + \infty )\)
LG c
\(\sqrt {2{x^2} - 1} > 1 - x\)
Phương pháp giải:
Biến đổi tương đương
\(\sqrt f > g \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
g < 0\\
f \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
g \ge 0\\
f > {g^2}
\end{array} \right.
\end{array} \right.\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sqrt {2{x^2} - 1} > 1 - x \cr &\Leftrightarrow \left[ \matrix{
\left\{ \matrix{
1 - x < 0 \hfill \cr
2{x^2} - 1 > 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
1 - x \ge 0 \hfill \cr
2{x^2} - 1 > {(1 - x)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
&\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 1\\{x^2} > \frac{1}{2}\left( {dung\,khi\,x > 1} \right)\end{array} \right.\\\left\{ \begin{array}{l}x \le 1\\2{x^2} - 1 > {x^2} - 2x + 1\end{array} \right.\end{array} \right.\cr &\Leftrightarrow \left[ \matrix{x > 1 \hfill \cr \left\{ \matrix{x \le 1 \hfill \cr {x^2} + 2x - 2 > 0 \hfill \cr} \right. \hfill \cr} \right. \cr &\Leftrightarrow \left[ \matrix{x > 1 \hfill \cr \left\{ \matrix{x \le 1 \hfill \cr \left[ \matrix{x < - 1 - \sqrt 3 \hfill \cr x > - 1 + \sqrt 3 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr} \right.\cr&\Leftrightarrow \left[ \begin{array}{l}x > 1\\\left\{ \begin{array}{l}x \le 1\\x < - 1 - \sqrt 3 \end{array} \right.\\\left\{ \begin{array}{l}x \le 1\\x > - 1 + \sqrt 3 \end{array} \right.\end{array} \right. \cr &\Leftrightarrow \left[ \begin{array}{l}x > 1\\x < - 1 - \sqrt 3 \\- 1 + \sqrt 3 < x \le 1\end{array} \right. \cr &\Leftrightarrow \left[ \begin{array}{l}x < - 1 - \sqrt 3 \\\left[ \begin{array}{l}x > 1\\- 1 + \sqrt 3 < x \le 1\end{array} \right.\end{array} \right. \cr &\Leftrightarrow \left[ \matrix{x < - 1 - \sqrt 3 \hfill \cr x > - 1 + \sqrt 3 \hfill \cr} \right. \cr} \)
Vậy \(S = ( - \infty , - 1 - \sqrt 3 ) \cup ( - 1 + \sqrt 3 , + \infty )\)
LG d
\(\sqrt {{x^2} - 5x - 14} \ge 2x - 1\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sqrt {{x^2} - 5x - 14} \ge 2x - 1 \cr
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
2x - 1 < 0 \hfill \cr
{x^2} - 5x - 14 \ge 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
2x - 1 \ge 0 \hfill \cr
{x^2} - 5x - 14 \ge {(2x - 1)^2} \hfill \cr} \right. \hfill \cr} \right.\cr& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x < {1 \over 2} \hfill \cr
\left[ \matrix{
x \le - 2 \hfill \cr
x \ge 7 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x \ge {1 \over 2} \hfill \cr
3{x^2} + x + 15 \le 0(VN) \hfill \cr} \right. \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \begin{array}{l}x < \frac{1}{2}\\\left[ \begin{array}{l}x \le - 2\\x \ge 7\end{array} \right.\end{array} \right.\cr} \) \(\Leftrightarrow x\le -2\).
Vậy \(S = (-∞, -2]\)
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