Whakaoti i te -3(x-9)(2+x)>0

$$-3(x-9)(2+x) > 0$$

Whakaoti mō x

$$x\in \left(-2,9\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x-9.

$$\left(-3x+27\right)\left(2+x\right)>0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -3x+27 ki te 2+x ka whakakotahi i ngā kupu rite.

$$21x-3x^{2}+54>0$$

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i 21x-3x^{2}+54. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$-21x+3x^{2}-54<0$$

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.

$$-21x+3x^{2}-54=0$$

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 3 mō te a, te -21 mō te b, me te -54 mō te c i te ture pūrua.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\left(-54\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{21±33}{6}$$

Whakaotia te whārite x=\frac{21±33}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$x=9 x=-2$$

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

$$3\left(x-9\right)\left(x+2\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-9 me te x+2. Whakaarohia te tauira ina he tōrunga te x-9 he tōraro te x+2.

$$x-9>0 x+2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+2 he tōraro te x-9.

$$x+2>0 x-9<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-2,9\right).

$$x\in \left(-2,9\right)$$

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

$$x\in \left(-2,9\right)$$