Whakaoti i te (x-3)^2<2

$$( x - 3 ) ^ { 2 } < 2$$

Whakaoti mō x

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$

Ngā Upane Otinga

Kia whakaotia te koreōrite, me tauwehe te taha mauī. Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$\left(x-3\right)^{2}=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -6 mō te b, me te 7 mō te c i te ture pūrua.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 7}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{6±2\sqrt{2}}{2}$$

Whakaotia te whārite x=\frac{6±2\sqrt{2}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\sqrt{2}+3 x=3-\sqrt{2}$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$\left(x-\left(\sqrt{2}+3\right)\right)\left(x-\left(3-\sqrt{2}\right)\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\left(\sqrt{2}+3\right) me te x-\left(3-\sqrt{2}\right). Whakaarohia te tauira ina he tōrunga te x-\left(\sqrt{2}+3\right) he tōraro te x-\left(3-\sqrt{2}\right).

$$x-\left(\sqrt{2}+3\right)>0 x-\left(3-\sqrt{2}\right)<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\left(3-\sqrt{2}\right) he tōraro te x-\left(\sqrt{2}+3\right).

$$x-\left(3-\sqrt{2}\right)>0 x-\left(\sqrt{2}+3\right)<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(3-\sqrt{2},\sqrt{2}+3\right).

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$

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

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$