Whakaoti i te 10/(x-5)(x+1)+x/x+1=3/x-5

$$\frac { 10 } { ( x - 5 ) ( x + 1 ) } + \frac { x } { x + 1 } = \frac { 3 } { x - 5 }$$

Whakaoti mō x

$$x=1$$

$$x=7$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,5 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(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o \left(x-5\right)\left(x+1\right),x+1,x-5.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x-5 ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$10+x^{2}-5x=3x+3$$

Tangohia te 3x mai i ngā taha e rua.

$$10+x^{2}-5x-3x=3$$

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

$$10+x^{2}-8x=3$$

Tangohia te 3 mai i ngā taha e rua.

$$10+x^{2}-8x-3=0$$

Tangohia te 3 i te 10, ka 7.

$$7+x^{2}-8x=0$$

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.

$$x^{2}-8x+7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -8 mō b, me 7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7}}{2}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 7}}{2}$$

Whakareatia -4 ki te 7.

$$x=\frac{-\left(-8\right)±\sqrt{64-28}}{2}$$

Tāpiri 64 ki te -28.

$$x=\frac{-\left(-8\right)±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-\left(-8\right)±6}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±6}{2}$$

Nā, me whakaoti te whārite x=\frac{8±6}{2} ina he tāpiri te ±. Tāpiri 8 ki te 6.

$$x=\frac{14}{2}$$

Whakawehe 14 ki te 2.

$$x=7$$

Nā, me whakaoti te whārite x=\frac{8±6}{2} ina he tango te ±. Tango 6 mai i 8.

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

Whakawehe 2 ki te 2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=7 x=1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,5 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(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o \left(x-5\right)\left(x+1\right),x+1,x-5.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x-5 ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$10+x^{2}-5x=3x+3$$

Tangohia te 3x mai i ngā taha e rua.

$$10+x^{2}-5x-3x=3$$

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

$$10+x^{2}-8x=3$$

Tangohia te 10 mai i ngā taha e rua.

$$x^{2}-8x=3-10$$

Tangohia te 10 i te 3, ka -7.

$$x^{2}-8x=-7$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=-7+16$$

Tāpiri -7 ki te 16.

$$x^{2}-8x+16=9$$

Tauwehea x^{2}-8x+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}.

$$\left(x-4\right)^{2}=9$$

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{9}$$

Whakarūnātia.

$$x-4=3 x-4=-3$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=7 x=1$$