Whakaoti i te x-1/x+1=2x-4/x+4

$$\frac{ x-1 }{ x+1 } = \frac{ 2x-4 }{ x+4 }$$

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 -4,-1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x+1\right)\left(x+4\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+1,x+4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+4 ki te x-1 ka whakakotahi i ngā kupu rite.

$$x^{2}+3x-4=\left(x+1\right)\left(2x-4\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-4 ka whakakotahi i ngā kupu rite.

$$x^{2}+3x-4=2x^{2}-2x-4$$

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

$$x^{2}+3x-4-2x^{2}=-2x-4$$

Pahekotia te x^{2} me -2x^{2}, ka -x^{2}.

$$-x^{2}+3x-4=-2x-4$$

Me tāpiri te 2x ki ngā taha e rua.

$$-x^{2}+3x-4+2x=-4$$

Pahekotia te 3x me 2x, ka 5x.

$$-x^{2}+5x-4=-4$$

Me tāpiri te 4 ki ngā taha e rua.

$$-x^{2}+5x-4+4=0$$

Tāpirihia te -4 ki te 4, 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}.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 5^{2}.

$$x=\frac{-5±5}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-5±5}{-2}$$

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

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

Whakawehe 0 ki te -2.

$$x=0$$

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

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

Whakawehe -10 ki te -2.

$$x=5$$

Kua oti te whārite te whakatau.

$$x=0 x=5$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+4 ki te x-1 ka whakakotahi i ngā kupu rite.

$$x^{2}+3x-4=\left(x+1\right)\left(2x-4\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-4 ka whakakotahi i ngā kupu rite.

$$x^{2}+3x-4=2x^{2}-2x-4$$

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

$$x^{2}+3x-4-2x^{2}=-2x-4$$

Pahekotia te x^{2} me -2x^{2}, ka -x^{2}.

$$-x^{2}+3x-4=-2x-4$$

Me tāpiri te 2x ki ngā taha e rua.

$$-x^{2}+3x-4+2x=-4$$

Pahekotia te 3x me 2x, ka 5x.

$$-x^{2}+5x-4=-4$$

Me tāpiri te 4 ki ngā taha e rua.

$$-x^{2}+5x=-4+4$$

Tāpirihia te -4 ki te 4, ka 0.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe 5 ki te -1.

$$x^{2}-5x=\frac{0}{-1}$$

Whakawehe 0 ki te -1.

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

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.

$$x=5 x=0$$