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Giải các bất phương trình:
a) \(\sqrt {{x^2} + x – 6} < x – 1\)
b) \(\sqrt {2x – 1} \le 2x – 3\)
c) \(\sqrt {2{x^2} – 1} > 1 – x\)
d) \(\sqrt {{x^2} – 5x – 14} \ge 2x – 1\)
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt {{x^2} + x – 6} < x – 1\cr& \Leftrightarrow \left\{ \matrix{
{x^2} + x – 6 \ge 0 \hfill \cr
x – 1 > 0 \hfill \cr
{x^2} + x – 6 < {(x – 1)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x \le 3 \hfill \cr
x \ge 2 \hfill \cr} \right. \hfill \cr
x > 1 \hfill \cr
3x < 7 \hfill \cr} \right. \Leftrightarrow 2 \le x < {7 \over 3} \cr} \)
Vậy \(S = {\rm{[}}2,{7 \over 3})\)
b) Ta có:
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\(\eqalign{
& \sqrt {2x – 1} \le 2x – 3 \Leftrightarrow \left\{ \matrix{
2x – 1 \ge 0 \hfill \cr
2x – 3 \ge 0 \hfill \cr
2x – 1 \le {(2x – 3)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \ge {1 \over 2} \hfill \cr
x \ge {3 \over 2} \hfill \cr
4{x^2} – 14x + 10 \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{
x \ge {3 \over 2} \hfill \cr
\left[ \matrix{
x \le 1 \hfill \cr
x \ge {5 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {5 \over 2} \cr} \)
Vậy \(S = {\rm{[}}{5 \over 2}; + \infty )\)
c) Ta có:
\(\eqalign{
& \sqrt {2{x^2} – 1} > 1 – x \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
1 – x < 0 \hfill \cr
2{x^2} – 1 > 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
1 – x \ge 0 \hfill \cr
2{x^2} – 1 > {(1 – x)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x > 1 \hfill \cr
\left\{ \matrix{
x \le 1 \hfill \cr
{x^2} + 2x – 2 > 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x > 1 \hfill \cr
\left\{ \matrix{
x \le 1 \hfill \cr
\left[ \matrix{
x < – 1 – \sqrt 3 \hfill \cr
x > – 1 + \sqrt 3 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr} \right.\cr& \Leftrightarrow \left[ \matrix{
x < – 1 – \sqrt 3 \hfill \cr
x > – 1 + \sqrt 3 \hfill \cr} \right. \cr} \)
Vậy \(S = ( – \infty , – 1 – \sqrt 3 ) \cup ( – 1 + \sqrt 3 , + \infty )\)
d) Ta có:
\(\eqalign{
& \sqrt {{x^2} – 5x – 14} \ge 2x – 1 \cr
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
2x – 1 < 0 \hfill \cr
{x^2} – 5x – 14 \ge 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
2x – 1 \ge 0 \hfill \cr
{x^2} – 5x – 14 \ge {(2x – 1)^2} \hfill \cr} \right. \hfill \cr} \right.\cr& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x < {1 \over 2} \hfill \cr
\left[ \matrix{
x \le – 2 \hfill \cr
x \ge 7 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x \ge {1 \over 2} \hfill \cr
3{x^2} + x + 15 \le 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \le – 2 \cr} \)
Vậy \(S = (-∞, -2]\)