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

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

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Hei kimi i te tauaro o x-1, kimihia te tauaro o ia taurangi.

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

Pahekotia te -x me -2x, ka -3x.

$$-3x+1+x^{2}+1>0$$

Tāpirihia te 1 ki te 1, ka 2.

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

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.

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

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 1 mō te a, te -3 mō te b, me te 2 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{3±1}{2}$$

Whakaotia te whārite x=\frac{3±1}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=1$$

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

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

Kia tōrunga te otinga, me tōraro tahi te x-2 me te x-1, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-2 me te x-1.

$$x-2<0 x-1<0$$

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

$$x<1$$

Whakaarohia te tauira ina he tōrunga tahi te x-2 me te x-1.

$$x-1>0 x-2>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>2.

$$x>2$$

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

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