Whakaoti i te -x^2+2x-1<0

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

Whakaoti mō x

$$x\neq 1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

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.

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

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

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

Mahia ngā tātaitai.

$$x=\frac{2±0}{2}$$

He ōrite ngā whakatau.

$$x=1$$

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

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

E mau ana te koreōrite mō x\neq 1.

$$x\neq 1$$