Whakaoti i te x=(2x-3)*(2x+3)/4x^2-16x+15

$$x= \frac{ (2x-3) \times (2x+3) }{ 4 { x }^{ 2 } -16x+15 }$$

Whakaoti mō x

$$x = \frac{\sqrt{73} + 7}{4} \approx 3.886000936$$

$$x=\frac{7-\sqrt{73}}{4}\approx -0.386000936$$

Ngā Upane Otinga

Whakaarohia te \left(2x-3\right)\left(2x+3\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 3.

$$x=\frac{\left(2x\right)^{2}-9}{4x^{2}-16x+15}$$

Whakarohaina te \left(2x\right)^{2}.

$$x=\frac{2^{2}x^{2}-9}{4x^{2}-16x+15}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$x=\frac{4x^{2}-9}{4x^{2}-16x+15}$$

Tangohia te \frac{4x^{2}-9}{4x^{2}-16x+15} mai i ngā taha e rua.

$$x-\frac{4x^{2}-9}{4x^{2}-16x+15}=0$$

Tauwehea te 4x^{2}-16x+15.

$$x-\frac{4x^{2}-9}{\left(2x-5\right)\left(2x-3\right)}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{\left(2x-5\right)\left(2x-3\right)}{\left(2x-5\right)\left(2x-3\right)}.

$$\frac{x\left(2x-5\right)\left(2x-3\right)}{\left(2x-5\right)\left(2x-3\right)}-\frac{4x^{2}-9}{\left(2x-5\right)\left(2x-3\right)}=0$$

Tā te mea he rite te tauraro o \frac{x\left(2x-5\right)\left(2x-3\right)}{\left(2x-5\right)\left(2x-3\right)} me \frac{4x^{2}-9}{\left(2x-5\right)\left(2x-3\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(2x-5\right)\left(2x-3\right)-\left(4x^{2}-9\right)}{\left(2x-5\right)\left(2x-3\right)}=0$$

Mahia ngā whakarea i roto o x\left(2x-5\right)\left(2x-3\right)-\left(4x^{2}-9\right).

$$\frac{4x^{3}-6x^{2}-10x^{2}+15x-4x^{2}+9}{\left(2x-5\right)\left(2x-3\right)}=0$$

Whakakotahitia ngā kupu rite i 4x^{3}-6x^{2}-10x^{2}+15x-4x^{2}+9.

$$\frac{4x^{3}-20x^{2}+15x+9}{\left(2x-5\right)\left(2x-3\right)}=0$$

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara \frac{3}{2},\frac{5}{2} nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(2x-5\right)\left(2x-3\right).

$$4x^{3}-20x^{2}+15x+9=0$$

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 9, ā, ka wehea e q te whakarea arahanga 4. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{3}{2}$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 4x^{3}-20x^{2}+15x+9 ki te 2\left(x-\frac{3}{2}\right)=2x-3, kia riro ko 2x^{2}-7x-3. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{2}-7x-3=0$$

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 2 mō te a, te -7 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{7±\sqrt{73}}{4}$$

Whakaotia te whārite 2x^{2}-7x-3=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{7-\sqrt{73}}{4} x=\frac{\sqrt{73}+7}{4}$$

Tangohia ngā uara e kore e ōrite ki te taurangi.

$$x\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$x=\frac{3}{2} x=\frac{7-\sqrt{73}}{4} x=\frac{\sqrt{73}+7}{4}$$

Tē taea kia ōrite te tāupe x ki \frac{3}{2}.

$$x=\frac{\sqrt{73}+7}{4} x=\frac{7-\sqrt{73}}{4}$$