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

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

Whakaoti mō x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakareatia te x+2 ki te x+2, ka \left(x+2\right)^{2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x^{2}-2.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te 3x+2 ka whakakotahi i ngā kupu rite.

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

Pahekotia te -2x^{2} me 3x^{2}, ka x^{2}.

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

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

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

Tangohia te 2 i te 4, ka 2.

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

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

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

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

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

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

Hei kimi i te tauaro o x^{2}+4x+4, kimihia te tauaro o ia taurangi.

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

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

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

Pahekotia te -4x me -4x, ka -8x.

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

Tangohia te 4 i te 4, ka 0.

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

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

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

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

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

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

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

Pahekotia te x^{2} me 2x^{2}, ka 3x^{2}.

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

Me tāpiri te 8x ki ngā taha e rua.

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

Pahekotia te -3x me 8x, ka 5x.

$$5x+3x^{2}+2=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$3x^{2}+5x+2=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=3\times 2=6$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 6.

$$1,6 2,3$$

Tātaihia te tapeke mō ia takirua.

$$1+6=7 2+3=5$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=2 b=3$$

Tuhia anō te 3x^{2}+5x+2 hei \left(3x^{2}+2x\right)+\left(3x+2\right).

$$\left(3x^{2}+2x\right)+\left(3x+2\right)$$

Whakatauwehea atu x i te 3x^{2}+2x.

$$x\left(3x+2\right)+3x+2$$

Whakatauwehea atu te kīanga pātahi 3x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x+2\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 3x+2=0 me te x+1=0.

$$x=-\frac{2}{3} x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakareatia te x+2 ki te x+2, ka \left(x+2\right)^{2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x^{2}-2.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te 3x+2 ka whakakotahi i ngā kupu rite.

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

Pahekotia te -2x^{2} me 3x^{2}, ka x^{2}.

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

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

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

Tangohia te 2 i te 4, ka 2.

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

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

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

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

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

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

Hei kimi i te tauaro o x^{2}+4x+4, kimihia te tauaro o ia taurangi.

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

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

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

Pahekotia te -4x me -4x, ka -8x.

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

Tangohia te 4 i te 4, ka 0.

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

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

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

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

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

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

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

Pahekotia te x^{2} me 2x^{2}, ka 3x^{2}.

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

Me tāpiri te 8x ki ngā taha e rua.

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

Pahekotia te -3x me 8x, ka 5x.

$$5x+3x^{2}+2=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$3x^{2}+5x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 5 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\times 2}}{2\times 3}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 3\times 2}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-5±\sqrt{25-12\times 2}}{2\times 3}$$

Whakareatia -12 ki te 2.

$$x=\frac{-5±\sqrt{25-24}}{2\times 3}$$

Tāpiri 25 ki te -24.

$$x=\frac{-5±\sqrt{1}}{2\times 3}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-5±1}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-5±1}{6}$$

Nā, me whakaoti te whārite x=\frac{-5±1}{6} ina he tāpiri te ±. Tāpiri -5 ki te 1.

$$x=-\frac{4}{6}$$

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

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

Nā, me whakaoti te whārite x=\frac{-5±1}{6} ina he tango te ±. Tango 1 mai i -5.

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

Whakawehe -6 ki te 6.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=-\frac{2}{3} x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakareatia te x+2 ki te x+2, ka \left(x+2\right)^{2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x^{2}-2.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te 3x+2 ka whakakotahi i ngā kupu rite.

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

Pahekotia te -2x^{2} me 3x^{2}, ka x^{2}.

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

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

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

Tangohia te 2 i te 4, ka 2.

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

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

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

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

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2\right)^{2}.

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

Hei kimi i te tauaro o x^{2}+4x+4, kimihia te tauaro o ia taurangi.

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

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

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

Pahekotia te -4x me -4x, ka -8x.

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

Tangohia te 4 i te 4, ka 0.

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

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

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

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

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

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

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

Pahekotia te x^{2} me 2x^{2}, ka 3x^{2}.

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

Me tāpiri te 8x ki ngā taha e rua.

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

Pahekotia te -3x me 8x, ka 5x.

$$5x+3x^{2}+2=0$$

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$5x+3x^{2}=-2$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$3x^{2}+5x=-2$$

Whakawehea ngā taha e rua ki te 3.

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

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

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

Whakawehea te \frac{5}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{6}. Nā, tāpiria te pūrua o te \frac{5}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{5}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{2}{3}+\frac{25}{36}$$

Tāpiri -\frac{2}{3} ki te \frac{25}{36} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{1}{36}$$

Tauwehea x^{2}+\frac{5}{3}x+\frac{25}{36}. 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}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{1}{36}$$

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

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Whakarūnātia.

$$x+\frac{5}{6}=\frac{1}{6} x+\frac{5}{6}=-\frac{1}{6}$$

Me tango \frac{5}{6} mai i ngā taha e rua o te whārite.

$$x=-\frac{2}{3} x=-1$$