Giải các hệ bất phương trình
a)
\(\left\{ \matrix{
2{x^2} + 9x + 7 > 0 \hfill \cr
{x^2} + x - 6 < 0 \hfill \cr} \right.\)
b)
\(\left\{ \matrix{
4{x^2} - 5x - 6 \le 0 \hfill \cr
- 4{x^2} + 12x - 5 < 0 \hfill \cr} \right.\)
c)
\(\left\{ \matrix{
- 2{x^2} - 5x + 4 \le 0 \hfill \cr
- {x^2} - 3x + 10 \ge 0 \hfill \cr} \right.\)
d)
\(\left\{ \matrix{
2{x^2} + x - 6 > 0 \hfill \cr
3{x^2} - 10x + 3 > 0 \hfill \cr} \right.\)
Đáp án
a) Ta có:
\(\eqalign{
& 2{x^2} + 9x + 7 > 0 \Leftrightarrow \left[ \matrix{
x < - {7 \over 2} \hfill \cr
x > - 1 \hfill \cr} \right. \cr
& {x^2} + x - 6 < 0 \Leftrightarrow - 3 < x < 2 \cr} \)
Do đó:
\(\left\{ \matrix{
2{x^2} + 9x + 7 > 0 \hfill \cr
{x^2} + x - 6 < 0 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < - {7 \over 2} \hfill \cr
x > - 1 \hfill \cr} \right. \hfill \cr
- 3 < x < 2 \hfill \cr} \right. \Leftrightarrow - 1 < x < 2\)
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Vậy tập nghiêm của hệ là \(S = (-1, 2)\)
b) Ta có:
\(\left\{ \matrix{
4{x^2} - 5x - 6 \le 0 \hfill \cr
- 4{x^2} + 12x - 5 < 0 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
- {3 \over 4} \le x \le 2 \hfill \cr
\left[ \matrix{
x < {1 \over 2} \hfill \cr
x > {5 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow - {3 \over 4} \le x < {1 \over 2}\)
Vậy tập nghiệm của hệ là \(S = {\rm{[}} - {3 \over 4};{1 \over 2}{\rm{]}}\)
c) Ta có:
\(\eqalign{
& \left\{ \matrix{
- 2{x^2} - 5x + 4 \le 0 \hfill \cr
- {x^2} - 3x + 10 \ge 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
2{x^2} + 5x - 4 \ge 0 \hfill \cr
{x^2} + 3x - 10 \le 0 \hfill \cr} \right. \cr
& \left\{ \matrix{
\left[ \matrix{
x \le {{ - 5 - \sqrt {57} } \over 4} \hfill \cr
x \ge {{ - 5 + \sqrt {57} } \over 4} \hfill \cr} \right. \hfill \cr
- 5 \le x \le 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
- 5 \le x \le {{ - 5 - \sqrt {57} } \over 4} \hfill \cr
{{ - 5 + \sqrt {57} } \over 4} \le x \le 2 \hfill \cr} \right. \cr} \)
Vậy \(S = {\rm{[}} - 5,{{ - 5 - \sqrt {57} } \over 4}{\rm{]}} \cup {\rm{[}}{{ - 5 + \sqrt {57} } \over 4};2{\rm{]}}\)
d) Ta có:
\(\left\{ \matrix{
2{x^2} + x - 6 > 0 \hfill \cr
3{x^2} - 10x + 3 > 0 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < - 2 \hfill \cr
x > {3 \over 2} \hfill \cr} \right. \hfill \cr
\left[ \matrix{
x < {1 \over 3} \hfill \cr
x > 3 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x < - 2 \hfill \cr
x > 3 \hfill \cr} \right.\)
Vậy \(S = ( - \infty , - 2) \cup (3, + \infty )\)