Whakaoti i te 2x+1/x+4x/2x+1=5

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

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,2x+1.

$$\left(2x+1\right)\left(2x+1\right)+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te 2x+1 ki te 2x+1, ka \left(2x+1\right)^{2}.

$$\left(2x+1\right)^{2}+x\times 4x=5x\left(2x+1\right)$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te x ki te x, ka x^{2}.

$$4x^{2}+4x+1+x^{2}\times 4=5x\left(2x+1\right)$$

Pahekotia te 4x^{2} me x^{2}\times 4, ka 8x^{2}.

$$8x^{2}+4x+1=5x\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 2x+1.

$$8x^{2}+4x+1=10x^{2}+5x$$

Tangohia te 10x^{2} mai i ngā taha e rua.

$$8x^{2}+4x+1-10x^{2}=5x$$

Pahekotia te 8x^{2} me -10x^{2}, ka -2x^{2}.

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

Tangohia te 5x mai i ngā taha e rua.

$$-2x^{2}+4x+1-5x=0$$

Pahekotia te 4x me -5x, ka -x.

$$-2x^{2}-x+1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -2x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-2=-2$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=-2$$

Tuhia anō te -2x^{2}-x+1 hei \left(-2x^{2}+x\right)+\left(-2x+1\right).

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

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

$$-x\left(2x-1\right)-\left(2x-1\right)$$

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-1\right)\left(-x-1\right)$$

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,2x+1.

$$\left(2x+1\right)\left(2x+1\right)+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te 2x+1 ki te 2x+1, ka \left(2x+1\right)^{2}.

$$\left(2x+1\right)^{2}+x\times 4x=5x\left(2x+1\right)$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te x ki te x, ka x^{2}.

$$4x^{2}+4x+1+x^{2}\times 4=5x\left(2x+1\right)$$

Pahekotia te 4x^{2} me x^{2}\times 4, ka 8x^{2}.

$$8x^{2}+4x+1=5x\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 2x+1.

$$8x^{2}+4x+1=10x^{2}+5x$$

Tangohia te 10x^{2} mai i ngā taha e rua.

$$8x^{2}+4x+1-10x^{2}=5x$$

Pahekotia te 8x^{2} me -10x^{2}, ka -2x^{2}.

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

Tangohia te 5x mai i ngā taha e rua.

$$-2x^{2}+4x+1-5x=0$$

Pahekotia te 4x me -5x, ka -x.

$$-2x^{2}-x+1=0$$

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

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

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\left(-2\right)}$$

Tāpiri 1 ki te 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-1\right)±3}{2\left(-2\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±3}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{1±3}{-4}$$

Nā, me whakaoti te whārite x=\frac{1±3}{-4} ina he tāpiri te ±. Tāpiri 1 ki te 3.

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

Whakawehe 4 ki te -4.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{1±3}{-4} ina he tango te ±. Tango 3 mai i 1.

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

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

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,2x+1.

$$\left(2x+1\right)\left(2x+1\right)+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te 2x+1 ki te 2x+1, ka \left(2x+1\right)^{2}.

$$\left(2x+1\right)^{2}+x\times 4x=5x\left(2x+1\right)$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1+x\times 4x=5x\left(2x+1\right)$$

Whakareatia te x ki te x, ka x^{2}.

$$4x^{2}+4x+1+x^{2}\times 4=5x\left(2x+1\right)$$

Pahekotia te 4x^{2} me x^{2}\times 4, ka 8x^{2}.

$$8x^{2}+4x+1=5x\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 2x+1.

$$8x^{2}+4x+1=10x^{2}+5x$$

Tangohia te 10x^{2} mai i ngā taha e rua.

$$8x^{2}+4x+1-10x^{2}=5x$$

Pahekotia te 8x^{2} me -10x^{2}, ka -2x^{2}.

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

Tangohia te 5x mai i ngā taha e rua.

$$-2x^{2}+4x+1-5x=0$$

Pahekotia te 4x me -5x, ka -x.

$$-2x^{2}-x+1=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea ngā taha e rua ki te -2.

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

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

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

Whakawehe -1 ki te -2.

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

Whakawehe -1 ki te -2.

$$x^{2}+\frac{1}{2}x=\frac{1}{2}$$

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.

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

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Tāpiri \frac{1}{2} ki te \frac{1}{16} 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Tauwehea x^{2}+\frac{1}{2}x+\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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

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

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

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