Whakaoti i te (x-2)*(x-3)*(x-4)=(x-2)*(x-3)*(x-5)

$$(x-2) \times (x-3) \times (x-4)=(x-2) \times (x-3) \times (x-5)$$

Whakaoti mō x

$$x=2$$

$$x=3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x-3 ka whakakotahi i ngā kupu rite.

$$\left(x^{2}-5x+6\right)\left(x-4\right)=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}-5x+6 ki te x-4 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x-3 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=\left(x^{2}-5x+6\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}-5x+6 ki te x-5 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=x^{3}-10x^{2}+31x-30$$

Tangohia te x^{3} mai i ngā taha e rua.

$$x^{3}-9x^{2}+26x-24-x^{3}=-10x^{2}+31x-30$$

Pahekotia te x^{3} me -x^{3}, ka 0.

$$-9x^{2}+26x-24=-10x^{2}+31x-30$$

Me tāpiri te 10x^{2} ki ngā taha e rua.

$$-9x^{2}+26x-24+10x^{2}=31x-30$$

Pahekotia te -9x^{2} me 10x^{2}, ka x^{2}.

$$x^{2}+26x-24=31x-30$$

Tangohia te 31x mai i ngā taha e rua.

$$x^{2}+26x-24-31x=-30$$

Pahekotia te 26x me -31x, ka -5x.

$$x^{2}-5x-24=-30$$

Me tāpiri te 30 ki ngā taha e rua.

$$x^{2}-5x-24+30=0$$

Tāpirihia te -24 ki te 30, ka 6.

$$x^{2}-5x+6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Tāpiri 25 ki te -24.

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

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±1}{2}$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tāpiri te ±. Tāpiri 5 ki te 1.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tango te ±. Tango 1 mai i 5.

$$x=\frac{4}{2}$$

Whakawehe 4 ki te 2.

$$x=2$$

Kua oti te whārite te whakatau.

$$x=3 x=2$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x-3 ka whakakotahi i ngā kupu rite.

$$\left(x^{2}-5x+6\right)\left(x-4\right)=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}-5x+6 ki te x-4 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-2 ki te x-3 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=\left(x^{2}-5x+6\right)\left(x-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}-5x+6 ki te x-5 ka whakakotahi i ngā kupu rite.

$$x^{3}-9x^{2}+26x-24=x^{3}-10x^{2}+31x-30$$

Tangohia te x^{3} mai i ngā taha e rua.

$$x^{3}-9x^{2}+26x-24-x^{3}=-10x^{2}+31x-30$$

Pahekotia te x^{3} me -x^{3}, ka 0.

$$-9x^{2}+26x-24=-10x^{2}+31x-30$$

Me tāpiri te 10x^{2} ki ngā taha e rua.

$$-9x^{2}+26x-24+10x^{2}=31x-30$$

Pahekotia te -9x^{2} me 10x^{2}, ka x^{2}.

$$x^{2}+26x-24=31x-30$$

Tangohia te 31x mai i ngā taha e rua.

$$x^{2}+26x-24-31x=-30$$

Pahekotia te 26x me -31x, ka -5x.

$$x^{2}-5x-24=-30$$

Me tāpiri te 24 ki ngā taha e rua.

$$x^{2}-5x=-30+24$$

Tāpirihia te -30 ki te 24, ka -6.

$$x^{2}-5x=-6$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Tāpiri -6 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

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

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

$$x=3 x=2$$