Bài 69 trang 154 SGK Đại số 10 nâng cao

a) \(|{{{x^2} - 2} \over {x + 1}}|\, = 2\)

b) \(|{{3x + 4} \over {x - 2}}|\, \le   3\)

c) \(|{{2x - 3} \over {x - 3}}|\,\, \ge 1\)

d) \(|2x + 3| = |4 – 3x|\)

Đáp án

a) Điều kiện: x ≠ - 1

Ta có:

\(\eqalign{
& |{{{x^2} - 2} \over {x + 1}}|\, = 2 \Leftrightarrow \left[ \matrix{
{{{x^2} - 2} \over {x + 1}} = 2 \hfill \cr
{{{x^2} - 2} \over {x + 1}} = - 2 \hfill \cr} \right. \cr&\Leftrightarrow \left[ \matrix{
{x^2} - 2 = 2x + 2 \hfill \cr
{x^2} - 2 = - 2x - 2 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{x^2} - 2x - 4 = 0 \hfill \cr
{x^2} + 2x = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 1 \pm \sqrt 5 \hfill \cr
\left[ \matrix{
x = 0 \hfill \cr
x = - 2 \hfill \cr} \right. \hfill \cr} \right. \cr} \)

Vậy \(S = {\rm{\{ }}1 \pm \sqrt 5 ;\,0;\,2\} \)

b) Điều kiện: x ≠  2

Ta có:

\(\eqalign{
& |{{3x + 4} \over {x - 2}}|\, \le  3 \Leftrightarrow |3x + 4|\, \le \,3|x - 2| \cr
& \Leftrightarrow {(3x + 4)^2} - 9{(x - 2)^2} \le 0 \cr
& \Leftrightarrow 10(6x - 2) \le 0 \Leftrightarrow x \le {1 \over 3} \cr} \)

Vậy \(S = ( - \infty ,{1 \over 3}{\rm{]}}\)

c) Điều kiện: x ≠ 3

Ta có:

\(\eqalign{
& |{{2x - 3} \over {x - 3}}|\,\, \ge 1\, \Leftrightarrow \,|2x - 3|\, \ge \,|x - 3| \cr
& \Leftrightarrow {(2x - 3)^2} - {(x - 3)^2} \ge 0 \cr
& \Leftrightarrow x(3x - 6) \ge 0 \Leftrightarrow \left[ \matrix{
x \le 0 \hfill \cr
x \ge 2 \hfill \cr} \right. \cr} \)

Vậy \(S = (-∞, 0] ∪ [2, 3) ∪ [3, +∞)\)

d) Ta có:

\(|2x + 3|\, = \,|4 - 3x|\, \Leftrightarrow \left[ \matrix{
2x + 3 = 4 - 3x \hfill \cr
2x + 3 = 3x - 4 \hfill \cr} \right. \)

\(\Leftrightarrow \left[ \matrix{
x = {1 \over 5} \hfill \cr
x = 7 \hfill \cr} \right.\)

Vậy \(S = {\rm{\{ }}{1 \over 5},7\} \)