Advertisements (Quảng cáo)
Giải các bất phương trình sau:
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:
\(\eqalign{
& {{ – 2{x^2} + 7x + 7} \over {{x^2} – 3x – 10}} \le – 1\cr& \Leftrightarrow {{ – 2{x^2} + 7x + 7} \over {{x^2} – 3x – 10}} + 1 \le 0 \Leftrightarrow {{ – {x^2} + 4x – 3} \over {{x^2} – 3x – 10}} \le 0 \cr} \)
Advertisements (Quảng cáo)
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 \(S = {\rm{[}} – \sqrt 3 ,\,\sqrt 3 {\rm{]}}\)