Whakaoti i te x+1/x-3=-x-6x+1/(x-3)(x-1)

$$\frac { x + 1 } { x - 3 } = - \frac { x - 6 x + 1 } { ( x - 3 ) ( x - 1 ) }$$

Whakaoti mō x

$$x=5$$

$$x=0$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$\left(x-1\right)\left(x+1\right)=-\left(x-6x+1\right)$$

Whakaarohia te \left(x-1\right)\left(x+1\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 1.

$$x^{2}-1=-\left(x-6x+1\right)$$

Pahekotia te x me -6x, ka -5x.

$$x^{2}-1=-\left(-5x+1\right)$$

Hei kimi i te tauaro o -5x+1, kimihia te tauaro o ia taurangi.

$$x^{2}-1=5x-1$$

Tangohia te 5x mai i ngā taha e rua.

$$x^{2}-1-5x=-1$$

Me tāpiri te 1 ki ngā taha e rua.

$$x^{2}-1-5x+1=0$$

Tāpirihia te -1 ki te 1, ka 0.

$$x^{2}-5x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Tuhia te pūtakerua o te \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±5}{2} ina he tāpiri te ±. Tāpiri 5 ki te 5.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{5±5}{2} ina he tango te ±. Tango 5 mai i 5.

$$x=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=5 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$\left(x-1\right)\left(x+1\right)=-\left(x-6x+1\right)$$

Whakaarohia te \left(x-1\right)\left(x+1\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 1.

$$x^{2}-1=-\left(x-6x+1\right)$$

Pahekotia te x me -6x, ka -5x.

$$x^{2}-1=-\left(-5x+1\right)$$

Hei kimi i te tauaro o -5x+1, kimihia te tauaro o ia taurangi.

$$x^{2}-1=5x-1$$

Tangohia te 5x mai i ngā taha e rua.

$$x^{2}-1-5x=-1$$

Me tāpiri te 1 ki ngā taha e rua.

$$x^{2}-5x=-1+1$$

Tāpirihia te -1 ki te 1, ka 0.

$$x^{2}-5x=0$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Tauwehea x^{2}-5x+\frac{25}{4}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=5 x=0$$