Whakaoti i te 12x^2geq10

$$12 x ^ { 2 } \geq 10$$

Whakaoti mō x

$$x\in (-\infty,-\frac{\sqrt{30}}{6}]\cup [\frac{\sqrt{30}}{6},\infty)$$

Ngā Upane Otinga

Whakawehea ngā taha e rua ki te 12. I te mea he tōrunga te 12, kāore e huri te ahunga koreōrite.

$$x^{2}\geq \frac{10}{12}$$

Whakahekea te hautanga \frac{10}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}\geq \frac{5}{6}$$

Tātaitia te pūtakerua o \frac{5}{6} kia tae ki \frac{\sqrt{30}}{6}. Tuhia anō te \frac{5}{6} hei \left(\frac{\sqrt{30}}{6}\right)^{2}.

$$x^{2}\geq \left(\frac{\sqrt{30}}{6}\right)^{2}$$

E mau ana te koreōrite mō |x|\geq \frac{\sqrt{30}}{6}.

$$|x|\geq \frac{\sqrt{30}}{6}$$

Tuhia anō te |x|\geq \frac{\sqrt{30}}{6} hei x\leq -\frac{\sqrt{30}}{6}\text{; }x\geq \frac{\sqrt{30}}{6}.

$$x\leq -\frac{\sqrt{30}}{6}\text{; }x\geq \frac{\sqrt{30}}{6}$$