Монгол Бодлогын Сан

Заавар: Эхлээд илэрхийллээ хялбарчил.

Бодолт: $$\dfrac{3x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2x^{-\frac{1}{3}}}=\dfrac{3x^{-\frac13}}{x^{-\frac13}(x-2)}=\dfrac{3}{x-2},$$ $$\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}=\dfrac{x^{\frac13}}{x^{\frac13}(x-1)}=\dfrac{1}{x-1}$$ тул $$\left(\displaystyle\frac{3x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2x^{-\frac{1}{3}}}-\displaystyle\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}\right)^{-1}=\left(\dfrac{3}{x-2}-\dfrac{1}{x-1}\right)^{-1}=$$ $$=\left(\dfrac{3x-3-x+2}{(x-1)(x-2)}\right)^{-1}=\left(\dfrac{2x-1}{x^2-3x+2}\right)^{-1}=\dfrac{x^2-3x+2}{2x-1}$$ байна. Иймд $$\left(\displaystyle\frac{3x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2x^{-\frac{1}{3}}}-\displaystyle\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}\right)^{-1}-\displaystyle\frac{3x-2}{1-2x}=$$ $$=\dfrac{x^2-3x+2}{2x-1}+\dfrac{3x-2}{2x-1}=\dfrac{x^2}{2x-1}$$ байна. $x=2+\sqrt2$ үед $$\dfrac{(2+\sqrt2)^2}{2(2+\sqrt2)-1}=\dfrac{4+4\sqrt2+2}{3+2\sqrt2}=2$$ байна.