[ĐẠI SỐ 10] Giải bất phương trình

V

vipboycodon

$x+\dfrac{x}{\sqrt{x^2-1}} > \dfrac{35}{12}$

$\leftrightarrow x^2+\dfrac{x^2}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}} > (\dfrac{35}{12})^2$

$\leftrightarrow \dfrac{x^4}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}} > \dfrac{1225}{144}$

$\leftrightarrow (\dfrac{x^2}{\sqrt{x^2-1}}+1)^2 > \dfrac{1369}{144}$

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