Bài 70 trang 154 SGK Đại số 10 nâng cao

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 \(S = (-∞, -2] ∪ [1, + ∞)\)