Whakaoti i te 2x+1/x-2+4/x=-8/x^2-2x

$$\frac { 2 x + 1 } { x - 2 } + \frac { 4 } { x } = \frac { - 8 } { x ^ { 2 } - 2 x }$$

Whakaoti mō x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Ngā Upane Whakamahi Tauwehe

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

$$x\left(2x+1\right)+\left(x-2\right)\times 4=-8$$

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

$$2x^{2}+x+\left(x-2\right)\times 4=-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te 4.

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

Pahekotia te x me 4x, ka 5x.

$$2x^{2}+5x-8=-8$$

Me tāpiri te 8 ki ngā taha e rua.

$$2x^{2}+5x-8+8=0$$

Tāpirihia te -8 ki te 8, ka 0.

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

Tauwehea te x.

$$x\left(2x+5\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2x+5=0.

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

Tē taea kia ōrite te tāupe x ki 0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$x\left(2x+1\right)+\left(x-2\right)\times 4=-8$$

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

$$2x^{2}+x+\left(x-2\right)\times 4=-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te 4.

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

Pahekotia te x me 4x, ka 5x.

$$2x^{2}+5x-8=-8$$

Me tāpiri te 8 ki ngā taha e rua.

$$2x^{2}+5x-8+8=0$$

Tāpirihia te -8 ki te 8, ka 0.

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

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

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

Tuhia te pūtakerua o te 5^{2}.

$$x=\frac{-5±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-5±5}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±5}{4} ina he tāpiri te ±. Tāpiri -5 ki te 5.

$$x=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-5±5}{4} ina he tango te ±. Tango 5 mai i -5.

$$x=-\frac{10}{4}$$

Whakahekea te hautanga \frac{-10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Tē taea kia ōrite te tāupe x ki 0.

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

Ngā Upane mō te Whakaoti i te Pūrua

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

$$x\left(2x+1\right)+\left(x-2\right)\times 4=-8$$

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

$$2x^{2}+x+\left(x-2\right)\times 4=-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te 4.

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

Pahekotia te x me 4x, ka 5x.

$$2x^{2}+5x-8=-8$$

Me tāpiri te 8 ki ngā taha e rua.

$$2x^{2}+5x=-8+8$$

Tāpirihia te -8 ki te 8, ka 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 0 ki te 2.

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

Whakawehea te \frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{4}. Nā, tāpiria te pūrua o te \frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

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

$$\left(x+\frac{5}{4}\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$x+\frac{5}{4}=\frac{5}{4} x+\frac{5}{4}=-\frac{5}{4}$$

Me tango \frac{5}{4} mai i ngā taha e rua o te whārite.

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

Tē taea kia ōrite te tāupe x ki 0.

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