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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:
\(\left\{ \matrix{
2{x^2} + 9x + 7 > 0 \hfill \cr
{x^2} + x – 6 < 0 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < – {7 \over 2} \hfill \cr
x > – 1 \hfill \cr} \right. \hfill \cr
– 3 < x < 2 \hfill \cr} \right. \Leftrightarrow – 1 < x < 2\)
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 \(S = ( – \infty , – 2) \cup (3, + \infty )\)