Whakaoti i te 1-t^3/1-t=7/5

$$\frac { 1 - t ^ { 3 } } { 1 - t } = \frac { 7 } { 5 }$$

Whakaoti mō t

$$t=\frac{\sqrt{65}}{10}-\frac{1}{2}\approx 0.306225775$$

$$t=-\frac{\sqrt{65}}{10}-\frac{1}{2}\approx -1.306225775$$

Ngā Upane Otinga

Tē taea kia ōrite te tāupe t ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 5\left(t-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 1-t,5.

$$-5\left(1-t^{3}\right)=7\left(t-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5 ki te 1-t^{3}.

$$-5+5t^{3}=7\left(t-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 7 ki te t-1.

$$-5+5t^{3}=7t-7$$

Tangohia te 7t mai i ngā taha e rua.

$$-5+5t^{3}-7t=-7$$

Me tāpiri te 7 ki ngā taha e rua.

$$-5+5t^{3}-7t+7=0$$

Tāpirihia te -5 ki te 7, ka 2.

$$2+5t^{3}-7t=0$$

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

$$5t^{3}-7t+2=0$$

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 2, ā, ka wehea e q te whakarea arahanga 5. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{2}{5},±2,±\frac{1}{5},±1$$

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.

$$t=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te t-k o te pūrau mō ia pūtake k. Whakawehea te 5t^{3}-7t+2 ki te t-1, kia riro ko 5t^{2}+5t-2. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$5t^{2}+5t-2=0$$

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 5 mō te a, te 5 mō te b, me te -2 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$t=\frac{-5±\sqrt{65}}{10}$$

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

$$t=-\frac{\sqrt{65}}{10}-\frac{1}{2} t=\frac{\sqrt{65}}{10}-\frac{1}{2}$$

Tangohia ngā uara e kore e ōrite ki te taurangi.

$$t\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$t=1 t=-\frac{\sqrt{65}}{10}-\frac{1}{2} t=\frac{\sqrt{65}}{10}-\frac{1}{2}$$

Tē taea kia ōrite te tāupe t ki 1.

$$t=\frac{\sqrt{65}}{10}-\frac{1}{2} t=-\frac{\sqrt{65}}{10}-\frac{1}{2}$$