Whakaoti i te -1/3(x+2)(x-1/3)>0

$$- \frac { 1 } { 3 } ( x + 2 ) ( x - \frac { 1 } { 3 } ) > 0$$

Whakaoti mō x

$$x\in \left(-2,\frac{1}{3}\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te -\frac{1}{3} ki te x+2.

$$\left(-\frac{1}{3}x-\frac{2}{3}\right)\left(x-\frac{1}{3}\right)>0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -\frac{1}{3}x-\frac{2}{3} ki te x-\frac{1}{3} ka whakakotahi i ngā kupu rite.

$$-\frac{1}{3}x^{2}-\frac{5}{9}x+\frac{2}{9}>0$$

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -\frac{1}{3}x^{2}-\frac{5}{9}x+\frac{2}{9}. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$\frac{1}{3}x^{2}+\frac{5}{9}x-\frac{2}{9}<0$$

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.

$$\frac{1}{3}x^{2}+\frac{5}{9}x-\frac{2}{9}=0$$

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 \frac{1}{3} mō te a, te \frac{5}{9} mō te b, me te -\frac{2}{9} mō te c i te ture pūrua.

$$x=\frac{-\frac{5}{9}±\sqrt{\left(\frac{5}{9}\right)^{2}-4\times \frac{1}{3}\left(-\frac{2}{9}\right)}}{\frac{1}{3}\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{-\frac{5}{9}±\frac{7}{9}}{\frac{2}{3}}$$

Whakaotia te whārite x=\frac{-\frac{5}{9}±\frac{7}{9}}{\frac{2}{3}} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{1}{3} x=-2$$

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

$$\frac{1}{3}\left(x-\frac{1}{3}\right)\left(x+2\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{1}{3} me te x+2. Whakaarohia te tauira ina he tōrunga te x-\frac{1}{3} he tōraro te x+2.

$$x-\frac{1}{3}>0 x+2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+2 he tōraro te x-\frac{1}{3}.

$$x+2>0 x-\frac{1}{3}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-2,\frac{1}{3}\right).

$$x\in \left(-2,\frac{1}{3}\right)$$

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

$$x\in \left(-2,\frac{1}{3}\right)$$