Whakaoti i te 2x^2>x

$$2 x ^ { 2 } > x$$

Whakaoti mō x

$$x\in \left(-\infty,0\right)\cup \left(\frac{1}{2},\infty\right)$$

Ngā Upane Otinga

Tangohia te x mai i ngā taha e rua.

$$2x^{2}-x>0$$

Tauwehea te x.

$$x\left(2x-1\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x me te x-\frac{1}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x me te x-\frac{1}{2}.

$$x<0 x-\frac{1}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<0.

$$x<0$$

Whakaarohia te tauira ina he tōrunga tahi te x me te x-\frac{1}{2}.

$$x-\frac{1}{2}>0 x>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{1}{2}.

$$x>\frac{1}{2}$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$x<0\text{; }x>\frac{1}{2}$$