Whakaoti i te (9-5x)^2+2(9-5x)^2-24<0

$$( 9 - 5 x ) ^ { 2 } + 2 ( 9 - 5 x ) ^ { 2 } - 24 < 0$$

Whakaoti mō x

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(9-5x\right)^{2}.

$$81-90x+25x^{2}+2\left(9-5x\right)^{2}-24<0$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(9-5x\right)^{2}.

$$81-90x+25x^{2}+2\left(81-90x+25x^{2}\right)-24<0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 81-90x+25x^{2}.

$$81-90x+25x^{2}+162-180x+50x^{2}-24<0$$

Tāpirihia te 81 ki te 162, ka 243.

$$243-90x+25x^{2}-180x+50x^{2}-24<0$$

Pahekotia te -90x me -180x, ka -270x.

$$243-270x+25x^{2}+50x^{2}-24<0$$

Pahekotia te 25x^{2} me 50x^{2}, ka 75x^{2}.

$$243-270x+75x^{2}-24<0$$

Tangohia te 24 i te 243, ka 219.

$$219-270x+75x^{2}<0$$

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.

$$219-270x+75x^{2}=0$$

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 75 mō te a, te -270 mō te b, me te 219 mō te c i te ture pūrua.

$$x=\frac{-\left(-270\right)±\sqrt{\left(-270\right)^{2}-4\times 75\times 219}}{2\times 75}$$

Mahia ngā tātaitai.

$$x=\frac{270±60\sqrt{2}}{150}$$

Whakaotia te whārite x=\frac{270±60\sqrt{2}}{150} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{2\sqrt{2}+9}{5} x=\frac{9-2\sqrt{2}}{5}$$

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

$$75\left(x-\frac{2\sqrt{2}+9}{5}\right)\left(x-\frac{9-2\sqrt{2}}{5}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{2\sqrt{2}+9}{5} me te x-\frac{9-2\sqrt{2}}{5}. Whakaarohia te tauira ina he tōrunga te x-\frac{2\sqrt{2}+9}{5} he tōraro te x-\frac{9-2\sqrt{2}}{5}.

$$x-\frac{2\sqrt{2}+9}{5}>0 x-\frac{9-2\sqrt{2}}{5}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\frac{9-2\sqrt{2}}{5} he tōraro te x-\frac{2\sqrt{2}+9}{5}.

$$x-\frac{9-2\sqrt{2}}{5}>0 x-\frac{2\sqrt{2}+9}{5}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right).

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

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

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$