Giải các bất phương trình
a) 3x2 - |5x + 2| >0
b) \(\sqrt {2{x^2} + 7x + 5} > x + 1\)
c) \(\sqrt {{x^2} + 4x - 5} \le x + 3\)
Đáp án
a) Ta có:
\(\eqalign{
& 3{x^2} - \left| {5x + 2} \right| > 0 \Leftrightarrow |5x + 2| < 3{x^2} \cr
& \Leftrightarrow - 3{x^2} < 5x + 2 < 3{x^2} \cr
& \Leftrightarrow \left\{ \matrix{
3{x^2} + 5x + 2 > 0 \hfill \cr
3{x^2} - 5x - 2 > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < - 1 \hfill \cr
x > - {2 \over 3} \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x < - {1 \over 3} \hfill \cr
x > 2 \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow \left[ \matrix{
x < - 1 \hfill \cr
- {2 \over 3} < x < - {1 \over 3} \hfill \cr
x > 2 \hfill \cr} \right. \cr} \)
Vậy: \(S = ( - \infty ,\, - 1) \cup ( - {2 \over 3}; - {1 \over 3}) \cup (2, + \infty )\)
b) Ta có:
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\(\eqalign{
& \sqrt {2{x^2} + 7x + 5} > x + 1 \cr
& \Leftrightarrow \,\,\left[ \matrix{
(I)\,\left\{ \matrix{
x + 1 < 0 \hfill \cr
2{x^2} + 7x + 5 \ge 0 \hfill \cr} \right. \hfill \cr
(II)\left\{ \matrix{
x + 1 \ge 0 \hfill \cr
2{x^2} + 7x + 5 > {(x + 1)^2} \hfill \cr} \right.\, \hfill \cr} \right. \cr} \)
Ta có:
\((I) \Leftrightarrow \left\{ \matrix{
x < - 1 \hfill \cr
\left[ \matrix{
x \le - {5 \over 2} \hfill \cr
x \ge - 1 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \le - {5 \over 2}\)
\((II) \Leftrightarrow \left\{ \matrix{
x \ge - 1 \hfill \cr
{x^2} + 5x + 4 > 0 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
x \ge - 1 \hfill \cr
\left[ \matrix{
x < - 4 \hfill \cr
x > - 1 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x > - 1\)
Vậy: \(S = ( - \infty ;\, - {5 \over 2}{\rm{]}}\, \cup ( - 1;\, + \infty )\)
c) Ta có:
\(\eqalign{
& \sqrt {{x^2} + 4x - 5} \le x + 3 \cr&\Leftrightarrow \left\{ \matrix{
x + 3 \ge 0 \hfill \cr
{x^2} + 4x - 5 \ge 0 \hfill \cr
{x^2} + 4x - 5 \le {(x + 3)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \ge - 3 \hfill \cr
\left[ \matrix{
x \le - 5 \hfill \cr
x \ge 1 \hfill \cr} \right. \hfill \cr
x \ge - 7 \hfill \cr} \right. \Leftrightarrow x \ge 1 \cr} \)
Vậy \(S = [1, +∞)\)