Whakaoti i te x^2-2x-1leq0

$$x ^ { 2 } - 2 x - 1 \leq 0$$

Whakaoti mō x

$$x\in \begin{bmatrix}1-\sqrt{2},\sqrt{2}+1\end{bmatrix}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

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

Mahia ngā tātaitai.

$$x=\frac{2±2\sqrt{2}}{2}$$

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

$$x=\sqrt{2}+1 x=1-\sqrt{2}$$

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

$$\left(x-\left(\sqrt{2}+1\right)\right)\left(x-\left(1-\sqrt{2}\right)\right)\leq 0$$

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-\left(\sqrt{2}+1\right) me x-\left(1-\sqrt{2}\right), me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x-\left(\sqrt{2}+1\right)\geq 0 me x-\left(1-\sqrt{2}\right)\leq 0.

$$x-\left(\sqrt{2}+1\right)\geq 0 x-\left(1-\sqrt{2}\right)\leq 0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina ko x-\left(\sqrt{2}+1\right)\leq 0 me x-\left(1-\sqrt{2}\right)\geq 0.

$$x-\left(1-\sqrt{2}\right)\geq 0 x-\left(\sqrt{2}+1\right)\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[1-\sqrt{2},\sqrt{2}+1\right].

$$x\in \begin{bmatrix}1-\sqrt{2},\sqrt{2}+1\end{bmatrix}$$

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

$$x\in \begin{bmatrix}1-\sqrt{2},\sqrt{2}+1\end{bmatrix}$$