Whakaoti i te b-2/b-1-5/b^2-4b+3+1=10/3-b

$$\frac { b - 2 } { b - 1 } - \frac { 5 } { b ^ { 2 } - 4 b + 3 } + 1 = \frac { 10 } { 3 - b }$$

Whakaoti mō b

$$b=-2$$

$$b = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe b 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(b-3\right)\left(b-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o b-1,b^{2}-4b+3,3-b.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Tangohia te 5 i te 6, ka 1.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tuaritanga hei whakarea te b-3 ki te b-1 ka whakakotahi i ngā kupu rite.

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

Pahekotia te b^{2} me b^{2}, ka 2b^{2}.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

Pahekotia te -5b me -4b, ka -9b.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

Tāpirihia te 1 ki te 3, ka 4.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1-b ki te 10.

$$2b^{2}-9b+4=10-10b$$

Tangohia te 10 mai i ngā taha e rua.

$$2b^{2}-9b+4-10=-10b$$

Tangohia te 10 i te 4, ka -6.

$$2b^{2}-9b-6=-10b$$

Me tāpiri te 10b ki ngā taha e rua.

$$2b^{2}-9b-6+10b=0$$

Pahekotia te -9b me 10b, ka b.

$$2b^{2}+b-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2b^{2}+ab+bb-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=2\left(-6\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$-1,12 -2,6 -3,4$$

Tātaihia te tapeke mō ia takirua.

$$-1+12=11 -2+6=4 -3+4=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-3 b=4$$

Tuhia anō te 2b^{2}+b-6 hei \left(2b^{2}-3b\right)+\left(4b-6\right).

$$\left(2b^{2}-3b\right)+\left(4b-6\right)$$

Tauwehea te b i te tuatahi me te 2 i te rōpū tuarua.

$$b\left(2b-3\right)+2\left(2b-3\right)$$

Whakatauwehea atu te kīanga pātahi 2b-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2b-3\right)\left(b+2\right)$$

Hei kimi otinga whārite, me whakaoti te 2b-3=0 me te b+2=0.

$$b=\frac{3}{2} b=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe b 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(b-3\right)\left(b-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o b-1,b^{2}-4b+3,3-b.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Tangohia te 5 i te 6, ka 1.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tuaritanga hei whakarea te b-3 ki te b-1 ka whakakotahi i ngā kupu rite.

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

Pahekotia te b^{2} me b^{2}, ka 2b^{2}.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

Pahekotia te -5b me -4b, ka -9b.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

Tāpirihia te 1 ki te 3, ka 4.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1-b ki te 10.

$$2b^{2}-9b+4=10-10b$$

Tangohia te 10 mai i ngā taha e rua.

$$2b^{2}-9b+4-10=-10b$$

Tangohia te 10 i te 4, ka -6.

$$2b^{2}-9b-6=-10b$$

Me tāpiri te 10b ki ngā taha e rua.

$$2b^{2}-9b-6+10b=0$$

Pahekotia te -9b me 10b, ka b.

$$2b^{2}+b-6=0$$

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

$$b=\frac{-1±\sqrt{1^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Pūrua 1.

$$b=\frac{-1±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$b=\frac{-1±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Whakareatia -8 ki te -6.

$$b=\frac{-1±\sqrt{1+48}}{2\times 2}$$

Tāpiri 1 ki te 48.

$$b=\frac{-1±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$b=\frac{-1±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$b=\frac{-1±7}{4}$$

Nā, me whakaoti te whārite b=\frac{-1±7}{4} ina he tāpiri te ±. Tāpiri -1 ki te 7.

$$b=\frac{6}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$b=\frac{3}{2}$$

Nā, me whakaoti te whārite b=\frac{-1±7}{4} ina he tango te ±. Tango 7 mai i -1.

$$b=-\frac{8}{4}$$

Whakawehe -8 ki te 4.

$$b=-2$$

Kua oti te whārite te whakatau.

$$b=\frac{3}{2} b=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe b 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(b-3\right)\left(b-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o b-1,b^{2}-4b+3,3-b.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Tangohia te 5 i te 6, ka 1.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tuaritanga hei whakarea te b-3 ki te b-1 ka whakakotahi i ngā kupu rite.

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

Pahekotia te b^{2} me b^{2}, ka 2b^{2}.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

Pahekotia te -5b me -4b, ka -9b.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

Tāpirihia te 1 ki te 3, ka 4.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1-b ki te 10.

$$2b^{2}-9b+4=10-10b$$

Me tāpiri te 10b ki ngā taha e rua.

$$2b^{2}-9b+4+10b=10$$

Pahekotia te -9b me 10b, ka b.

$$2b^{2}+b+4=10$$

Tangohia te 4 mai i ngā taha e rua.

$$2b^{2}+b=10-4$$

Tangohia te 4 i te 10, ka 6.

$$2b^{2}+b=6$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2b^{2}+b}{2}=\frac{6}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$b^{2}+\frac{1}{2}b=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$b^{2}+\frac{1}{2}b=3$$

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$b^{2}+\frac{1}{2}b+\left(\frac{1}{4}\right)^{2}=3+\left(\frac{1}{4}\right)^{2}$$

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$b^{2}+\frac{1}{2}b+\frac{1}{16}=3+\frac{1}{16}$$

Tāpiri 3 ki te \frac{1}{16}.

$$b^{2}+\frac{1}{2}b+\frac{1}{16}=\frac{49}{16}$$

Tauwehea b^{2}+\frac{1}{2}b+\frac{1}{16}. 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}.

$$\left(b+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

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

$$\sqrt{\left(b+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Whakarūnātia.

$$b+\frac{1}{4}=\frac{7}{4} b+\frac{1}{4}=-\frac{7}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$b=\frac{3}{2} b=-2$$