Bài 18 trang 223 SGK Đại số 10 Nâng cao

LG a

3x² - |5x + 2| >0

Lời giải chi tiết:

Ta có:

\(\eqalign{
& 3{x^2} - \left| {5x + 2} \right| > 0 \cr &\Leftrightarrow |5x + 2| < 3{x^2} \cr 
& \Leftrightarrow - 3{x^2} < 5x + 2 < 3{x^2} \cr 
& \Leftrightarrow \left\{ \matrix{
3{x^2} + 5x + 2 > 0 \hfill \cr 
3{x^2} - 5x - 2 > 0 \hfill \cr} \right. \cr &\Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x < - 1 \hfill \cr 
x > - {2 \over 3} \hfill \cr} \right. \hfill \cr 
\left\{ \matrix{
x < - {1 \over 3} \hfill \cr 
x > 2 \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow \left[ \matrix{
x < - 1 \hfill \cr 
- {2 \over 3} < x < - {1 \over 3} \hfill \cr 
x > 2 \hfill \cr} \right. \cr} \)

Vậy: \(S = ( - \infty ,\, - 1) \cup ( - {2 \over 3}; - {1 \over 3}) \cup (2, + \infty )\)

Cách khác:

LG b

\(\sqrt {2{x^2} + 7x + 5}  > x + 1\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {2{x^2} + 7x + 5} > x + 1 \cr 
& \Leftrightarrow \,\,\left[ \matrix{
(I)\,\left\{ \matrix{
x + 1 < 0 \hfill \cr 
2{x^2} + 7x + 5 \ge 0 \hfill \cr} \right. \hfill \cr 
(II)\left\{ \matrix{
x + 1 \ge 0 \hfill \cr 
2{x^2} + 7x + 5 > {(x + 1)^2} \hfill \cr} \right.\, \hfill \cr} \right. \cr} \) 

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: \(S = ( - \infty ;\, - {5 \over 2}{\rm{]}}\, \cup ( - 1;\, + \infty )\)

LG c

\(\sqrt {{x^2} + 4x - 5}  \le x + 3\)

Lời giải chi tiết:

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 \(S = [1, +∞)\)

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