Whakaoti i te left(x-1right)^3=0

$${ \left(x-1 \right) }^{ 3 } =0$$

Whakaoti mō x

$$x=1$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} hei whakaroha \left(x-1\right)^{3}.

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

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -1, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-3x^{2}+3x-1 ki te x-1, kia riro ko x^{2}-2x+1. Whakaotihia te whārite ina ōrite te hua ki te 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.

$$x=1$$