Whakaoti i te 2x(1-x)+1-x<0

$$2 x ( 1 - x ) + 1 - x < 0$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te 1-x.

$$2x-2x^{2}+1-x<0$$

Pahekotia te 2x me -x, ka x.

$$x-2x^{2}+1<0$$

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

$$-x+2x^{2}-1>0$$

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.

$$-x+2x^{2}-1=0$$

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 2 mō te a, te -1 mō te b, me te -1 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{1±3}{4}$$

Whakaotia te whārite x=\frac{1±3}{4} ina he tōrunga te ±, ina he tōraro te ±.

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

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

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

Kia tōrunga te otinga, me tōraro tahi te x-1 me te x+\frac{1}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-1 me te x+\frac{1}{2}.

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

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

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

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

Te otinga e whakaea i ngā koreōrite e rua ko x>1.

$$x>1$$

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

$$x<-\frac{1}{2}\text{; }x>1$$