Bài 72 trang 154 SGK Đại số 10 nâng cao

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:

\(\eqalign{
& {{2x - 4} \over {\sqrt {{x^2} - 3x - 10} }} > 1\cr& \Leftrightarrow \left\{ \matrix{
{x^2} - 3x - 10 > 0 \hfill \cr
\sqrt {{x^2} - 3x - 10} < 2x - 4 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{x^2} - 3x - 10 > 0 \hfill \cr
2x - 4 > 0 \hfill \cr
{x^2} - 3x - 10 < {(2x - 4)^2} \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < - 2 \hfill \cr
x > 5 \hfill \cr} \right. \hfill \cr
x > 2 \hfill \cr
3{x^2} - 13x + 26 > 0\,\,(\forall x) \hfill \cr} \right. \Leftrightarrow x > 5 \cr} \)

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 \(S = (-∞, 0] ∪ [34, +∞)\)