Câu 25 trang 121 SGK Đại số 10 nâng cao

a) \({{x + 2} \over 3} - x + 1 > x + 3\)

b) \({{3x + 5} \over 2} - 1 \le {{x + 2} \over 3} + x\)

c) \((1 - \sqrt 2 )x < 3 - 2\sqrt 2 \)

d) \({(x + \sqrt 3 )^2} \ge {(x - \sqrt 3 )^2} + 2\)

Đáp án 

a) Ta có:

\(\eqalign{
& {{x + 2} \over 3} - x + 1 > x + 3\cr& \Leftrightarrow x + 2 - 3x + 3 > 3x + 9 \cr
& \Leftrightarrow - 5x < 4 \Leftrightarrow x < - {4 \over 5} \cr} \)

Vậy  \(S = ( - \infty ; - {4 \over 5})\)  

b) Ta có:

\(\eqalign{
& {{3x + 5} \over 2} - 1 \le {{x + 2} \over 3} + x \cr&\Leftrightarrow 9x + 15 - 6 \le 2x + 4 + 6x \cr
& \Leftrightarrow x \le -5 \cr} \)

Vậy \(S = (-∞; -5)\)

c)

\(\eqalign{
& (1 - \sqrt 2 )x < 3 - 2\sqrt 2 \Leftrightarrow (1 - \sqrt 2 )x < {(1 - \sqrt 2 )^2} \cr
& \Leftrightarrow x > {{{{(1 - \sqrt 2 )}^2}} \over {1 - \sqrt 2 }} = 1 - \sqrt 2 \,\,(do\;1 - \sqrt 2 < 0) \cr} \) 

Vậy \(S = (1 - \sqrt 2 ; + \infty )\)

d)

\(\eqalign{
& {(x + \sqrt 3 )^2} \ge {(x - \sqrt 3 )^2} + 2 \cr
& \Leftrightarrow {(x + \sqrt 3 )^2} - {(x - \sqrt 3 )^2} \ge 2 \cr
& \Leftrightarrow 4\sqrt 3 x \ge 2 \Leftrightarrow x \ge {1 \over {2\sqrt 3 }} \cr} \)

Vậy \(S = {\rm{[}}{1 \over {2\sqrt 3 }};\, + \infty )\)