Nội dung bài giảng
Giải các bất phương trình sau
a) \(\sqrt {{x^2} - x - 12} \ge x - 1\)
b) \(\sqrt {{x^2} - 4x - 12} > 2x + 3\)
c) \({{\sqrt {x + 5} } \over {1 - x}} < 1\)
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt {{x^2} - x - 12} \ge x - 1\cr& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x - 1 < 0 \hfill \cr
{x^2} - x - 12 \ge 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x - 1 \ge 0 \hfill \cr
{x^2} - x - 12 \ge {(x - 1)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x < 1 \hfill \cr
\left[ \matrix{
x \le - 3 \hfill \cr
x \ge 4 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x \ge 1 \hfill \cr
x \ge 13 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x \le - 3 \hfill \cr
x \ge 13 \hfill \cr} \right. \cr} \)
Vậy \(S = (-∞, -3] ∪ [13, +∞)\)
b) Ta có:
\(\eqalign{
& \sqrt {{x^2} - 4x - 12} > 2x + 3 \cr&\Leftrightarrow \left[ \matrix{
\left\{ \matrix{
2x + 3 < 0 \hfill \cr
{x^2} - 4x - 12 \ge 0 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
2x - 3 \ge 0 \hfill \cr
{x^2} - 4x - 12 > {(2x + 3)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x < - {3 \over 2} \hfill \cr
\left[ \matrix{
x \le - 2 \hfill \cr
x \ge 6 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x \ge {3 \over 2} \hfill \cr
3{x^2} + 16x + 21 < 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x \le - 2 \hfill \cr
\left\{ \matrix{
x \ge {3 \over 2} \hfill \cr
- 3 < x < - {7 \over 3} \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow x < - 2 \cr} \)
Vậy \(S = (-∞, -2]\)
c) Bất phương trình đã cho tương đương với:
\((I)\,\left\{ \matrix{
1 - x > 0 \hfill \cr
\sqrt {x + 5} < 1 - x \hfill \cr} \right.\,\,\,\,;\,\,\,\,(II)\left\{ \matrix{
1 - x < 0 \hfill \cr
\sqrt {x + 5} > 1 - x \hfill \cr} \right.\)
\(\eqalign{
& (I) \Leftrightarrow \left\{ \matrix{
x < 1 \hfill \cr
x + 5 \ge 0 \hfill \cr
x + 5 < {(1 - x)^2} \hfill \cr
- 5 \le x < 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x < 1 \hfill \cr
x \ge - 5 \hfill \cr
{x^2} - 3x - 4 > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x < 1 \hfill \cr
x \ge - 5 \hfill \cr
{x^2} - 3x - 4 > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
- 5 \le x < 1 \hfill \cr
\left[ \matrix{
x < - 1 \hfill \cr
x > 4 \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow - 5 \le x < 1 \cr} \)
Vậy \(S = [-5, -1) ∪ (1, +∞)\)