Advertisements (Quảng cáo)
Giải các bất phương trình sau
a) \(\sqrt {{x^2} + 6x + 8} \le 2x + 3\)
b) \({{2x – 4} \over {\sqrt {{x^2} – 3x – 10} }} > 1\)
c) \(6\sqrt {(x – 2)(x – 32)} \le {x^2} – 34x + 48\)
Đáp án
a)
Áp dụng:
\(\sqrt A = B \Leftrightarrow \left\{ \matrix{
A \ge 0 \hfill \cr
B \ge 0 \hfill \cr
A \le {B^2} \hfill \cr} \right.\)
Ta có:
\(\eqalign{
& \sqrt {{x^2} + 6x + 8} \le 2x + 3 \cr&\Leftrightarrow \left\{ \matrix{
{x^2} + 6x + 8 \ge 0 \hfill \cr
2x + 3 \ge 0 \hfill \cr
{x^2} + 6x + 8 \le {(2x + 3)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x \le – 4 \hfill \cr
x \ge – 2 \hfill \cr} \right. \hfill \cr
x \ge – {3 \over 2} \hfill \cr
3{x^2} + 6x + 1 \ge 0 \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{
x \ge – {3 \over 2} \hfill \cr
\left[ \matrix{
x \le {{ – 3 – \sqrt 6 } \over 3} \hfill \cr
x \ge {{ – 3 + \sqrt 6 } \over 3} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {{\sqrt 6 } \over 3} – 1 \cr} \)
Vậy \(S = {\rm{[}}{{\sqrt 6 } \over 3} – 1, + \infty )\)
b) Ta có:
Advertisements (Quảng cáo)
\(\eqalign{
& {{2x – 4} \over {\sqrt {{x^2} – 3x – 10} }} > 1\cr& \Leftrightarrow \left\{ \matrix{
{x^2} – 3x – 10 > 0 \hfill \cr
\sqrt {{x^2} – 3x – 10} < 2x – 4 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{x^2} – 3x – 10 > 0 \hfill \cr
2x – 4 > 0 \hfill \cr
{x^2} – 3x – 10 < {(2x – 4)^2} \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < – 2 \hfill \cr
x > 5 \hfill \cr} \right. \hfill \cr
x > 2 \hfill \cr
3{x^2} – 13x + 26 > 0\,\,(\forall x) \hfill \cr} \right. \Leftrightarrow x > 5 \cr} \)
Vậy \(S = (5, +∞)\)
c) Đặt \(y = \sqrt {(x – 2)(x – 32)} = \sqrt {{x^2} – 34x + 64} \,\,\,(y \ge 0)\)
⇒ x2 – 34x = y2 – 64
Ta có bất phương trình:
6y ≤ y2 – 16 ⇔ y2 – 6y – 16 ≥ 0 ⇔ y ≤ 2 hoặc y ≥ 8
Với điều kiện y ≥ 0, ta có:
y ≥ 8 ⇔ x2 – 34x + 64 ≥ 64 ⇔ x2 – 34x ≥ 0
⇔ x ≤ 0 hoặc x ≥ 34
Vậy \(S = (-∞, 0] ∪ [34, +∞)\)