Whakaoti i te x-4/x+3=11/x^2+5x+6

$$\frac { x - 4 } { x + 3 } = 1 \frac { 1 } { x ^ { 2 } + 5 x + 6 }$$

Whakaoti mō x

$$x=\sqrt{10}+1\approx 4.16227766$$

$$x=1-\sqrt{10}\approx -2.16227766$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x+2\right)\left(x+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+3,x^{2}+5x+6.

$$\left(x+2\right)\left(x-4\right)=1\times 1$$

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

$$x^{2}-2x-8=1\times 1$$

Whakareatia te 1 ki te 1, ka 1.

$$x^{2}-2x-8=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-2x-8-1=0$$

Tangohia te 1 i te -8, ka -9.

$$x^{2}-2x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -2 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-9\right)}}{2}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-9\right)}}{2}$$

Whakareatia -4 ki te -9.

$$x=\frac{-\left(-2\right)±\sqrt{4+36}}{2}$$

Tāpiri 4 ki te 36.

$$x=\frac{-\left(-2\right)±\sqrt{40}}{2}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-\left(-2\right)±2\sqrt{10}}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{10}}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{10}}{2} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{10}.

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

Whakawehe 2+2\sqrt{10} ki te 2.

$$x=\sqrt{10}+1$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{10}}{2} ina he tango te ±. Tango 2\sqrt{10} mai i 2.

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

Whakawehe 2-2\sqrt{10} ki te 2.

$$x=1-\sqrt{10}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x+2\right)\left(x+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+3,x^{2}+5x+6.

$$\left(x+2\right)\left(x-4\right)=1\times 1$$

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

$$x^{2}-2x-8=1\times 1$$

Whakareatia te 1 ki te 1, ka 1.

$$x^{2}-2x-8=1$$

Me tāpiri te 8 ki ngā taha e rua.

$$x^{2}-2x=1+8$$

Tāpirihia te 1 ki te 8, ka 9.

$$x^{2}-2x=9$$

Whakawehea te -2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -1. Nā, tāpiria te pūrua o te -1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-2x+1=9+1$$

Tāpiri 9 ki te 1.

$$x^{2}-2x+1=10$$

Tauwehea x^{2}-2x+1. 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}.

$$\left(x-1\right)^{2}=10$$

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

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{10}$$

Whakarūnātia.

$$x-1=\sqrt{10} x-1=-\sqrt{10}$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$