Whakaoti i te 5x-5/2x+5=2x-11/x-5

$$\frac { 5 x - 5 } { 2 x + 5 } = \frac { 2 x - 11 } { x - 5 }$$

Whakaoti mō x

$$x=8$$

$$x=10$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{5}{2},5 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-5\right)\left(2x+5\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+5,x-5.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-5 ki te 5x-5 ka whakakotahi i ngā kupu rite.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+5 ki te 2x-11 ka whakakotahi i ngā kupu rite.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

Pahekotia te 5x^{2} me -4x^{2}, ka x^{2}.

$$x^{2}-30x+25=-12x-55$$

Me tāpiri te 12x ki ngā taha e rua.

$$x^{2}-30x+25+12x=-55$$

Pahekotia te -30x me 12x, ka -18x.

$$x^{2}-18x+25=-55$$

Me tāpiri te 55 ki ngā taha e rua.

$$x^{2}-18x+25+55=0$$

Tāpirihia te 25 ki te 55, ka 80.

$$x^{2}-18x+80=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -18 mō b, me 80 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 80}}{2}$$

Pūrua -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 80}}{2}$$

Whakareatia -4 ki te 80.

$$x=\frac{-\left(-18\right)±\sqrt{324-320}}{2}$$

Tāpiri 324 ki te -320.

$$x=\frac{-\left(-18\right)±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-18\right)±2}{2}$$

Ko te tauaro o -18 ko 18.

$$x=\frac{18±2}{2}$$

Nā, me whakaoti te whārite x=\frac{18±2}{2} ina he tāpiri te ±. Tāpiri 18 ki te 2.

$$x=\frac{20}{2}$$

Whakawehe 20 ki te 2.

$$x=10$$

Nā, me whakaoti te whārite x=\frac{18±2}{2} ina he tango te ±. Tango 2 mai i 18.

$$x=\frac{16}{2}$$

Whakawehe 16 ki te 2.

$$x=8$$

Kua oti te whārite te whakatau.

$$x=10 x=8$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{5}{2},5 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-5\right)\left(2x+5\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+5,x-5.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-5 ki te 5x-5 ka whakakotahi i ngā kupu rite.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+5 ki te 2x-11 ka whakakotahi i ngā kupu rite.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

Pahekotia te 5x^{2} me -4x^{2}, ka x^{2}.

$$x^{2}-30x+25=-12x-55$$

Me tāpiri te 12x ki ngā taha e rua.

$$x^{2}-30x+25+12x=-55$$

Pahekotia te -30x me 12x, ka -18x.

$$x^{2}-18x+25=-55$$

Tangohia te 25 mai i ngā taha e rua.

$$x^{2}-18x=-55-25$$

Tangohia te 25 i te -55, ka -80.

$$x^{2}-18x=-80$$

Whakawehea te -18, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -9. Nā, tāpiria te pūrua o te -9 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-18x+\left(-9\right)^{2}=-80+\left(-9\right)^{2}$$

Pūrua -9.

$$x^{2}-18x+81=-80+81$$

Tāpiri -80 ki te 81.

$$x^{2}-18x+81=1$$

Tauwehea x^{2}-18x+81. 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}.

$$\left(x-9\right)^{2}=1$$

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

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

Whakarūnātia.

$$x-9=1 x-9=-1$$

Me tāpiri 9 ki ngā taha e rua o te whārite.

$$x=10 x=8$$