Whakaoti i te n^3-n=6n^2-6

$$n ^ { 3 } - n = 6 n ^ { 2 } - 6$$

Whakaoti mō n

$$n=1$$

$$n=6$$

$$n=-1$$

Ngā Upane Otinga

Tangohia te 6n^{2} mai i ngā taha e rua.

$$n^{3}-n-6n^{2}=-6$$

Me tāpiri te 6 ki ngā taha e rua.

$$n^{3}-n-6n^{2}+6=0$$

Hurinahatia te whārite ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$n^{3}-6n^{2}-n+6=0$$

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 6, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$n=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te n-k o te pūrau mō ia pūtake k. Whakawehea te n^{3}-6n^{2}-n+6 ki te n-1, kia riro ko n^{2}-5n-6. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$n^{2}-5n-6=0$$

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 -5 mō te b, me te -6 mō te c i te ture pūrua.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Mahia ngā tātaitai.

$$n=\frac{5±7}{2}$$

Whakaotia te whārite n^{2}-5n-6=0 ina he tōrunga te ±, ina he tōraro te ±.

$$n=-1 n=6$$

Rārangitia ngā otinga katoa i kitea.

$$n=1 n=-1 n=6$$