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Giải các bất phương trình
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:
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\(\eqalign{
& {{3x + 5} \over 2} – 1 \le {{x + 2} \over 3} + x \cr&\Leftrightarrow 9x + 15 – 6 \le 2x + 4 + 6x \cr
& \Leftrightarrow x \le -5 \cr} \)
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 \(S = {\rm{[}}{1 \over {2\sqrt 3 }};\, + \infty )\)