H
huytrandinh
bài 2 ta có
$x^{4}+x^{2}+1=(x^{2}+x+1)(x^{2}-x+1)$
$t=\frac{x}{x^{2}+1} (|t|\leq \frac{1}{2})$
$=>bpt<=>\sqrt{(\frac{1}{t}-1)(t+1)}\geq \sqrt{1-t}-t=>-1\leq t\leq 1$
$<=>\sqrt{1-t}(\sqrt{\frac{t+1}{t}}-1)\geq -t$
$=>1\geq t> 0$
$<=>\sqrt{t-1}(\sqrt{t+1}-\sqrt{t})\geq -t.\sqrt{t}$
$VT\geq 0,VP< 0=>0\leq t\leq 1=>0< t\leq \frac{1}{2}$
$.t> 0<=>x> 0$
$.\frac{x}{x^{2}+1}\leq \frac{1}{2}<=>(x-1)^{2}\geq 0$
$=>S=(0,+\infty )$
$x^{4}+x^{2}+1=(x^{2}+x+1)(x^{2}-x+1)$
$t=\frac{x}{x^{2}+1} (|t|\leq \frac{1}{2})$
$=>bpt<=>\sqrt{(\frac{1}{t}-1)(t+1)}\geq \sqrt{1-t}-t=>-1\leq t\leq 1$
$<=>\sqrt{1-t}(\sqrt{\frac{t+1}{t}}-1)\geq -t$
$=>1\geq t> 0$
$<=>\sqrt{t-1}(\sqrt{t+1}-\sqrt{t})\geq -t.\sqrt{t}$
$VT\geq 0,VP< 0=>0\leq t\leq 1=>0< t\leq \frac{1}{2}$
$.t> 0<=>x> 0$
$.\frac{x}{x^{2}+1}\leq \frac{1}{2}<=>(x-1)^{2}\geq 0$
$=>S=(0,+\infty )$