Bài 62 trang 136 SGK Đại số 10 nâng cao

a) 

\(\left\{ \matrix{
4x - 3 < 3x + 4 \hfill \cr
{x^2} - 7x + 10 \le 0 \hfill \cr} \right.\)

b) 

\(\left\{ \matrix{
2{x^2} + 9x - 7 > 0 \hfill \cr
{x^2} + x - 6 \le 0 \hfill \cr} \right.\)

c)

\(\left\{ \matrix{
{x^2} - 9 < 0 \hfill \cr
(x - 1)(3{x^2} + 7x + 4) \ge 0 \hfill \cr} \right.\)

Đáp án

a) Ta có: 

\( \Leftrightarrow \left\{ \matrix{
4x - 3 < 3x + 4 \hfill \cr
{x^2} - 7x + 10 \le 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x < 7 \hfill \cr
2 \le x \le 5 \hfill \cr} \right.\)

\(\Leftrightarrow 2 \le x \le 5\)

Vậy \(S = [2, 5]\)

b) Ta có:

\(\eqalign{
& \left\{ \matrix{
2{x^2} + 9x - 7 > 0 \hfill \cr
{x^2} + x - 6 \le 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < {{ - 9 - \sqrt {137} } \over 4} \hfill \cr
x > {{ - 9 + \sqrt {137} } \over 4} \hfill \cr} \right. \hfill \cr
- 3 \le x \le 2 \hfill \cr} \right. \cr
& \Leftrightarrow {{ - 9 + \sqrt {137} } \over 4} < x < 2 \cr} \) 

Vậy \(S = ({{ - 9 + \sqrt {137} } \over 4};2{\rm{]}}\)

c) Ta có:

\(\eqalign{
& \left\{ \matrix{
{x^2} - 9 < 0 \hfill \cr
(x - 1)(3{x^2} + 7x + 4) \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{
- 3 < x < 3 \hfill \cr
\left[ \matrix{
- {4 \over 3} \le x \le - 1 \hfill \cr
x \ge 1 \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
- {4 \over 3} \le x \le - 1 \hfill \cr
1 \le x \le 3 \hfill \cr} \right. \cr} \)

Vậy \(S = \,{\rm{[}} - {4 \over 3},\, - 1{\rm{]}}\, \cup {\rm{[}}1,\,3)\)