Question

cho H = \(\dfrac{x^2}{x^4+x^2+1}\). Tính H biết \(\dfrac{x}{x^2+x+1}\) = \(\dfrac{2}{3}\)

Answer 1

Ta có \(\dfrac{x}{x^2+x+1}=\dfrac{2}{3}\Leftrightarrow3x=2x^2+2x+2\Leftrightarrow2x^2-x+2=0\Leftrightarrow\left(\sqrt{2}x\right)^2-2\sqrt{2}x.\dfrac{1}{2\sqrt{2}}+\dfrac{1}{8}+\dfrac{15}{8}=0\Leftrightarrow\left(\sqrt{2}x+\dfrac{1}{2\sqrt{2}}\right)^2+\dfrac{15}{8}=0\)Ta có \(\left(\sqrt{2}x+\dfrac{1}{2\sqrt{2}}\right)^2\ge0\Rightarrow\left(\sqrt{2}x+\dfrac{1}{2\sqrt{2}}\right)^2+\dfrac{15}{8}\ge\dfrac{15}{8}\)

Nên không có giá trị x

Vậy không có giá trị H