Whakaoti i te (3+7x-5/x^2-x-2-3x/x+1):4/x-2+5/x+3=frac{9x^2+43x+8}{4(x^2+4x+3)}

$$( 3 + \frac { 7 x - 5 } { x ^ { 2 } - x - 2 } - \frac { 3 x } { x + 1 } ) : \frac { 4 } { x - 2 } + \frac { 5 } { x + 3 } = \frac { 9 x ^ { 2 } + 43 x + 8 } { 4 ( x ^ { 2 } + 4 x + 3 ) }$$

Whakaoti mō x

$$x=7$$

Ngā Upane Otinga

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 4\left(x+1\right)\left(x+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+3,4\left(x^{2}+4x+3\right).

$$\left(x+1\right)\left(x+3\right)\left(x-2\right)\left(3+\frac{7x-5}{x^{2}-x-2}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+3 ka whakakotahi i ngā kupu rite.

$$\left(x^{2}+4x+3\right)\left(x-2\right)\left(3+\frac{7x-5}{x^{2}-x-2}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}+4x+3 ki te x-2 ka whakakotahi i ngā kupu rite.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(3+\frac{7x-5}{x^{2}-x-2}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Tauwehea te x^{2}-x-2.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(3+\frac{7x-5}{\left(x-2\right)\left(x+1\right)}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 3 ki te \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(\frac{3\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{7x-5}{\left(x-2\right)\left(x+1\right)}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Tā te mea he rite te tauraro o \frac{3\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} me \frac{7x-5}{\left(x-2\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(\frac{3\left(x-2\right)\left(x+1\right)+7x-5}{\left(x-2\right)\left(x+1\right)}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Mahia ngā whakarea i roto o 3\left(x-2\right)\left(x+1\right)+7x-5.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(\frac{3x^{2}+3x-6x-6+7x-5}{\left(x-2\right)\left(x+1\right)}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Whakakotahitia ngā kupu rite i 3x^{2}+3x-6x-6+7x-5.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(\frac{3x^{2}+4x-11}{\left(x-2\right)\left(x+1\right)}-\frac{3x}{x+1}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o \left(x-2\right)\left(x+1\right) me x+1 ko \left(x-2\right)\left(x+1\right). Whakareatia \frac{3x}{x+1} ki te \frac{x-2}{x-2}.

$$\left(x^{3}+2x^{2}-5x-6\right)\left(\frac{3x^{2}+4x-11}{\left(x-2\right)\left(x+1\right)}-\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}\right)+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Tā te mea he rite te tauraro o \frac{3x^{2}+4x-11}{\left(x-2\right)\left(x+1\right)} me \frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\left(x^{3}+2x^{2}-5x-6\right)\times \frac{3x^{2}+4x-11-3x\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Mahia ngā whakarea i roto o 3x^{2}+4x-11-3x\left(x-2\right).

$$\left(x^{3}+2x^{2}-5x-6\right)\times \frac{3x^{2}+4x-11-3x^{2}+6x}{\left(x-2\right)\left(x+1\right)}+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Whakakotahitia ngā kupu rite i 3x^{2}+4x-11-3x^{2}+6x.

$$\left(x^{3}+2x^{2}-5x-6\right)\times \frac{10x-11}{\left(x-2\right)\left(x+1\right)}+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Tuhia te \left(x^{3}+2x^{2}-5x-6\right)\times \frac{10x-11}{\left(x-2\right)\left(x+1\right)} hei hautanga kotahi.

$$\frac{\left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)}{\left(x-2\right)\left(x+1\right)}+\left(4x+4\right)\times 5=9x^{2}+43x+8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4x+4 ki te 5.

$$\frac{\left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)}{\left(x-2\right)\left(x+1\right)}+20x+20=9x^{2}+43x+8$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 20x+20 ki te \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\frac{\left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)}{\left(x-2\right)\left(x+1\right)}+\frac{\left(20x+20\right)\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=9x^{2}+43x+8$$

Tā te mea he rite te tauraro o \frac{\left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)}{\left(x-2\right)\left(x+1\right)} me \frac{\left(20x+20\right)\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{\left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)+\left(20x+20\right)\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=9x^{2}+43x+8$$

Mahia ngā whakarea i roto o \left(x^{3}+2x^{2}-5x-6\right)\left(10x-11\right)+\left(20x+20\right)\left(x-2\right)\left(x+1\right).

$$\frac{10x^{4}-11x^{3}+20x^{3}-22x^{2}-50x^{2}+55x-60x+66+20x^{3}-20x^{2}-40x+20x^{2}-20x-40}{\left(x-2\right)\left(x+1\right)}=9x^{2}+43x+8$$

Whakakotahitia ngā kupu rite i 10x^{4}-11x^{3}+20x^{3}-22x^{2}-50x^{2}+55x-60x+66+20x^{3}-20x^{2}-40x+20x^{2}-20x-40.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}=9x^{2}+43x+8$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{x^{2}-x-2}=9x^{2}+43x+8$$

Tangohia te 9x^{2} mai i ngā taha e rua.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{x^{2}-x-2}-9x^{2}=43x+8$$

Tauwehea te x^{2}-x-2.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}-9x^{2}=43x+8$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia -9x^{2} ki te \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}+\frac{-9x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=43x+8$$

Tā te mea he rite te tauraro o \frac{10x^{4}+29x^{3}-72x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)} me \frac{-9x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26-9x^{2}\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=43x+8$$

Mahia ngā whakarea i roto o 10x^{4}+29x^{3}-72x^{2}-65x+26-9x^{2}\left(x-2\right)\left(x+1\right).

$$\frac{10x^{4}+29x^{3}-72x^{2}-65x+26-9x^{4}-9x^{3}+18x^{3}+18x^{2}}{\left(x-2\right)\left(x+1\right)}=43x+8$$

Whakakotahitia ngā kupu rite i 10x^{4}+29x^{3}-72x^{2}-65x+26-9x^{4}-9x^{3}+18x^{3}+18x^{2}.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}=43x+8$$

Tangohia te 43x mai i ngā taha e rua.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}-43x=8$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26}{x^{2}-x-2}-43x=8$$

Tauwehea te x^{2}-x-2.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}-43x=8$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia -43x ki te \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)}+\frac{-43x\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=8$$

Tā te mea he rite te tauraro o \frac{x^{4}+38x^{3}-54x^{2}-65x+26}{\left(x-2\right)\left(x+1\right)} me \frac{-43x\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26-43x\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=8$$

Mahia ngā whakarea i roto o x^{4}+38x^{3}-54x^{2}-65x+26-43x\left(x-2\right)\left(x+1\right).

$$\frac{x^{4}+38x^{3}-54x^{2}-65x+26-43x^{3}-43x^{2}+86x^{2}+86x}{\left(x-2\right)\left(x+1\right)}=8$$

Whakakotahitia ngā kupu rite i x^{4}+38x^{3}-54x^{2}-65x+26-43x^{3}-43x^{2}+86x^{2}+86x.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26}{\left(x-2\right)\left(x+1\right)}=8$$

Tangohia te 8 mai i ngā taha e rua.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26}{\left(x-2\right)\left(x+1\right)}-8=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26}{x^{2}-x-2}-8=0$$

Tauwehea te x^{2}-x-2.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26}{\left(x-2\right)\left(x+1\right)}-8=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 8 ki te \frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26}{\left(x-2\right)\left(x+1\right)}-\frac{8\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=0$$

Tā te mea he rite te tauraro o \frac{x^{4}-5x^{3}-11x^{2}+21x+26}{\left(x-2\right)\left(x+1\right)} me \frac{8\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26-8\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=0$$

Mahia ngā whakarea i roto o x^{4}-5x^{3}-11x^{2}+21x+26-8\left(x-2\right)\left(x+1\right).

$$\frac{x^{4}-5x^{3}-11x^{2}+21x+26-8x^{2}-8x+16x+16}{\left(x-2\right)\left(x+1\right)}=0$$

Whakakotahitia ngā kupu rite i x^{4}-5x^{3}-11x^{2}+21x+26-8x^{2}-8x+16x+16.

$$\frac{x^{4}-5x^{3}-19x^{2}+29x+42}{\left(x-2\right)\left(x+1\right)}=0$$

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-2\right)\left(x+1\right).

$$x^{4}-5x^{3}-19x^{2}+29x+42=0$$

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

$$±42,±21,±14,±7,±6,±3,±2,±1$$

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.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{4}-5x^{3}-19x^{2}+29x+42 ki te x+1, kia riro ko x^{3}-6x^{2}-13x+42. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{3}-6x^{2}-13x+42=0$$

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

$$±42,±21,±14,±7,±6,±3,±2,±1$$

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.

$$x=2$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-6x^{2}-13x+42 ki te x-2, kia riro ko x^{2}-4x-21. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}-4x-21=0$$

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 -4 mō te b, me te -21 mō te c i te ture pūrua.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-21\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{4±10}{2}$$

Whakaotia te whārite x^{2}-4x-21=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-3 x=7$$

Tangohia ngā uara e kore e ōrite ki te taurangi.

$$x=7$$

Rārangitia ngā otinga katoa i kitea.

$$x=-1 x=2 x=-3 x=7$$

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,2,-3.

$$x=7$$