Giải các bất phương trình sau:
a) \(\left\{ \matrix{
{x^2} \ge 0,25 \hfill \cr
{x^2} – x \le 0 \hfill \cr} \right.;\)
b) \(\left\{ \matrix{
(x – 1)(2x + 3) > 0 \hfill \cr
(x – 4)(x + {1 \over 4}) \le 0 \hfill \cr} \right.\)
Gợi ý làm bài
a) \(\eqalign{
& \left\{ \matrix{
{x^2} \ge 4x \hfill \cr
{(2x – 1)^2} < 9 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
{x^2} – 4x \ge 0 \hfill \cr
– 3 < 2x – 1 < 3 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \in ( – \infty ;0] \cup {\rm{[}}4; + \infty ) \hfill \cr
– 1 < x < 2 \hfill \cr} \right. \Leftrightarrow – 1 < x \le 0 \cr} \)
b) \(\eqalign{
& \left\{ \matrix{
2x – 3 < (x + 1)(x – 2) \hfill \cr
{x^2} – x \le 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
{x^2} – 3x + 1 > 0 \hfill \cr
{x^2} – x – 6 \le 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \in ( – \infty ;{{3 – \sqrt 5 } \over 2}) \cup ({{3 + \sqrt 5 } \over 2};3] \hfill \cr
– 2 \le x \le 3 \hfill \cr} \right. \cr} \)
\( \Leftrightarrow x \in {\rm{[ – 2;}}{{3 – \sqrt 5 } \over 2}) \cup ({{3 + \sqrt 5 } \over 2};3{\rm{]}}\)