Whakaoti i te x-4/x+3=1x/x^2+5x+6

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

Whakaoti mō x

$$x = \frac{\sqrt{41} + 3}{2} \approx 4.701562119$$

$$x=\frac{3-\sqrt{41}}{2}\approx -1.701562119$$

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.

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

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

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

Tangohia te 1x mai i ngā taha e rua.

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

Pahekotia te -2x me -x, ka -3x.

$$x^{2}-3x-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-8\right)}}{2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-8\right)}}{2}$$

Whakareatia -4 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2}$$

Tāpiri 9 ki te 32.

$$x=\frac{-\left(-3\right)±\sqrt{41}}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{41}}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{41}}{2} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+3}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{41}}{2} ina he tango te ±. Tango \sqrt{41} mai i 3.

$$x=\frac{3-\sqrt{41}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}+3}{2} x=\frac{3-\sqrt{41}}{2}$$

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.

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

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

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

Tangohia te 1x mai i ngā taha e rua.

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

Pahekotia te -2x me -x, ka -3x.

$$x^{2}-3x-8=0$$

Me tāpiri te 8 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}-3x=8$$

Whakawehea te -3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{2}. Nā, tāpiria te pūrua o te -\frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=8+\left(-\frac{3}{2}\right)^{2}$$

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=8+\frac{9}{4}$$

Tāpiri 8 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{41}{4}$$

Tauwehea x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{41}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{\sqrt{41}}{2} x-\frac{3}{2}=-\frac{\sqrt{41}}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}+3}{2} x=\frac{3-\sqrt{41}}{2}$$