Bài 54 trang 135 SGK Đại số 10 nâng cao

a) \({{{x^2} - 9x + 14} \over {{x^2} - 5x + 4}} > 0\)

b) \({{ - 2{x^2} + 7x + 7} \over {{x^2} - 3x - 10}} \le  - 1\)

c) (2x + 1)(x² + x – 30) ≥ 0

d) x⁴ – 3x² ≤ 0

Đáp án

a) Ta có:

\(\eqalign{
& {x^2} - 9x + 14 = 0 \Leftrightarrow \left[ \matrix{
x = 2 \hfill \cr
x = 7 \hfill \cr} \right. \cr
& {x^2} - 5x + 4 = 0 \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
x = 4 \hfill \cr} \right. \cr} \)

Bảng xét dấu:

Vậy \(S = (-∞, 1) ∪ (2, 4) ∪ (7, +∞)\)

b) Ta có:

\(\eqalign{
& {{ - 2{x^2} + 7x + 7} \over {{x^2} - 3x - 10}} \le - 1\cr& \Leftrightarrow {{ - 2{x^2} + 7x + 7} \over {{x^2} - 3x - 10}} + 1 \le 0 \Leftrightarrow {{ - {x^2} + 4x - 3} \over {{x^2} - 3x - 10}} \le 0 \cr} \)

Ta lại có:

\(\eqalign{
& - {x^2} + 4x - 3 = 0 \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
x = 3 \hfill \cr} \right. \cr
& {x^2} - 3x - 10 = 0 \Leftrightarrow \left[ \matrix{
x = 5 \hfill \cr
x = - 2 \hfill \cr} \right. \cr} \) 

Bảng xét dấu:

Vậy \(S = (-∞, -2) ∪ [1, 3] ∪ (5, +∞)\)

c) Bảng xét dấu:

Vậy \(S = {\rm{[}} - 6,\, - {1 \over 2}{\rm{]}} \cup {\rm{[}}5,\, + \infty )\)

d) Ta có:

\(\eqalign{
& {x^4} - 3{x^2} \le 0 \Leftrightarrow {x^2}({x^2} - 3) \le 0 \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
{x^2} - 3 \le 0 \hfill \cr} \right. \cr
& \Leftrightarrow - \sqrt 3 \le x \le \sqrt 3 \cr} \)

Vậy \(S = {\rm{[}} - \sqrt 3 ,\,\sqrt 3 {\rm{]}}\)