Bài 67 trang 151 SGK Đại số 10 nâng cao

Giải các bất phương trình:

LG a

 \(\sqrt {{x^2} + x - 6}  < x - 1\)

Giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {{x^2} + x - 6} < x - 1\cr& \Leftrightarrow \left\{ \matrix{
{x^2} + x - 6 \ge 0 \hfill \cr 
x - 1 > 0 \hfill \cr 
{x^2} + x - 6 < {(x - 1)^2} \hfill \cr} \right. \cr 
& \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x \le 3 \hfill \cr 
x \ge 2 \hfill \cr} \right. \hfill \cr 
x > 1 \hfill \cr 
3x < 7 \hfill \cr} \right. \Leftrightarrow 2 \le x < {7 \over 3} \cr} \)

Vậy \(S = {\rm{[}}2,{7 \over 3})\)

LG b

\(\sqrt {2x - 1}  \le 2x - 3\)

Giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {2x - 1} \le 2x - 3 \Leftrightarrow \left\{ \matrix{
2x - 1 \ge 0 \hfill \cr 
2x - 3 \ge 0 \hfill \cr 
2x - 1 \le {(2x - 3)^2} \hfill \cr} \right. \cr 
& \Leftrightarrow \left\{ \matrix{
x \ge {1 \over 2} \hfill \cr 
x \ge {3 \over 2} \hfill \cr 
4{x^2} - 14x + 10 \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{
x \ge {3 \over 2} \hfill \cr 
\left[ \matrix{
x \le 1 \hfill \cr 
x \ge {5 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {5 \over 2} \cr} \) 

Vậy \(S = {\rm{[}}{5 \over 2}; + \infty )\)

LG c

\(\sqrt {2{x^2} - 1}  > 1 - x\)

Giải chi tiết:

Ta có: 

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

Vậy \(S = ( - \infty , - 1 - \sqrt 3 ) \cup ( - 1 + \sqrt 3 , + \infty )\)

LG d

\(\sqrt {{x^2} - 5x - 14}  \ge 2x - 1\)

Giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {{x^2} - 5x - 14} \ge 2x - 1 \cr 
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
2x - 1 < 0 \hfill \cr 
{x^2} - 5x - 14 \ge 0 \hfill \cr} \right. \hfill \cr 
\left\{ \matrix{
2x - 1 \ge 0 \hfill \cr 
{x^2} - 5x - 14 \ge {(2x - 1)^2} \hfill \cr} \right. \hfill \cr} \right.\cr& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x < {1 \over 2} \hfill \cr 
\left[ \matrix{
x \le - 2 \hfill \cr 
x \ge 7 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr 
\left\{ \matrix{
x \ge {1 \over 2} \hfill \cr 
3{x^2} + x + 15 \le 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \le - 2 \cr} \) 

Vậy \(S = (-∞, -2]\)