Nội dung bài giảng
Giải các bất phương trình sau:
a) |x2 – 5x + 4| ≤ x2 + 6x + 5
b) 4x2 + 4x - |2x + 1| ≥ 5
Đáp án
a) Áp dụng:
|A| ≤ B ⇔ -B ≤ A ≤ B
|x2 – 5x + 4| ≤ x2 + 6x + 5
⇔ -x2 – 6x – 5 ≤ x2 – 5x + 4 ≤ x2 + 6x + 5
\(\left\{ \matrix{
2{x^2} + x + 9 \ge 0 \hfill \cr
11x \ge - 1 \hfill \cr} \right. \Leftrightarrow x \ge - {1 \over {11}}\)
Vậy \(S = {\rm{[}} - {1 \over {11}}; + \infty )\)
b) Ta có: 4x2 + 4x - |2x + 1| ≥ 5
⇔ |2x + 1| ≤ 4x2 + 4x – 5
⇔ -4x2 – 4x + 5 ≤ 2x + 1 ≤ 4x2 + 4x – 5
\( \Leftrightarrow \left\{ \matrix{
4{x^2} + 6x - 4 \ge 0 \hfill \cr
4{x^2} + 2x - 6 \ge 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x \le - 2 \hfill \cr
x \ge {1 \over 2} \hfill \cr} \right. \hfill \cr
\left[ \matrix{
x \le - {3 \over 2} \hfill \cr
x \ge 1 \hfill \cr} \right. \hfill \cr} \right. \)
\(\Leftrightarrow \left[ \matrix{
x \le - 2 \hfill \cr
x \ge 1 \hfill \cr} \right.\)
Vậy \(S = (-∞, -2] ∪ [1, + ∞)\)