[Toán 10] Bất đẳng thức

T

tranvanhung7997

$A = x^4 - \sqrt[]{x^5} + x - \sqrt[]{x} + 1$ với $x \ge 0$
$= x^4 - x^2\sqrt[]{x} + x - \sqrt[]{x} + 1$
$= (x^4 - 2.x^2.\dfrac{1}{2}.\sqrt[]{x} + \dfrac{x}{4}) + \dfrac{3}{4}.(x - 2.\sqrt[]{x}.\dfrac{2}{3} + \dfrac{4}{9}) + \dfrac{2}{3}$
$= (x^2 - \dfrac{1}{2}.\sqrt[]{x})^2 + \dfrac{3}{4}.(\sqrt[]{x} - \dfrac{2}{3})^2 + \dfrac{2}{3} > 0$ với $ x \ge 0$
 
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