Whakaoti i te -f/10f+42=1/f+3

$$\frac { - f } { 10 f + 42 } = \frac { 1 } { f + 3 }$$

Whakaoti mō f

$$f=-7$$

$$f=-6$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe f ki tētahi o ngā uara -\frac{21}{5},-3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 2\left(f+3\right)\left(5f+21\right), arā, te tauraro pātahi he tino iti rawa te kitea o 10f+42,f+3.

$$\left(f+3\right)\left(-f\right)=10f+42$$

Whakamahia te āhuatanga tohatoha hei whakarea te f+3 ki te -f.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Tangohia te 10f mai i ngā taha e rua.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

Tangohia te 42 mai i ngā taha e rua.

$$f\left(-f\right)+3\left(-f\right)-10f-42=0$$

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

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f-42=0$$

Whakareatia te 3 ki te -1, ka -3.

$$f^{2}\left(-1\right)-3f-10f-42=0$$

Pahekotia te -3f me -10f, ka -13f.

$$f^{2}\left(-1\right)-13f-42=0$$

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.

$$-f^{2}-13f-42=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, -13 mō b, me -42 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$f=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

Pūrua -13.

$$f=\frac{-\left(-13\right)±\sqrt{169-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$f=\frac{-\left(-13\right)±\sqrt{169+4\left(-42\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -42.

$$f=\frac{-\left(-13\right)±\sqrt{169-168}}{2\left(-1\right)}$$

Tāpiri 169 ki te -168.

$$f=\frac{-\left(-13\right)±\sqrt{1}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 1.

$$f=\frac{-\left(-13\right)±1}{2\left(-1\right)}$$

Ko te tauaro o -13 ko 13.

$$f=\frac{13±1}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$f=\frac{13±1}{-2}$$

Nā, me whakaoti te whārite f=\frac{13±1}{-2} ina he tāpiri te ±. Tāpiri 13 ki te 1.

$$f=\frac{14}{-2}$$

Whakawehe 14 ki te -2.

$$f=-7$$

Nā, me whakaoti te whārite f=\frac{13±1}{-2} ina he tango te ±. Tango 1 mai i 13.

$$f=\frac{12}{-2}$$

Whakawehe 12 ki te -2.

$$f=-6$$

Kua oti te whārite te whakatau.

$$f=-7 f=-6$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe f ki tētahi o ngā uara -\frac{21}{5},-3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 2\left(f+3\right)\left(5f+21\right), arā, te tauraro pātahi he tino iti rawa te kitea o 10f+42,f+3.

$$\left(f+3\right)\left(-f\right)=10f+42$$

Whakamahia te āhuatanga tohatoha hei whakarea te f+3 ki te -f.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Tangohia te 10f mai i ngā taha e rua.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

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

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f=42$$

Whakareatia te 3 ki te -1, ka -3.

$$f^{2}\left(-1\right)-3f-10f=42$$

Pahekotia te -3f me -10f, ka -13f.

$$f^{2}\left(-1\right)-13f=42$$

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.

$$-f^{2}-13f=42$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-f^{2}-13f}{-1}=\frac{42}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$f^{2}+\left(-\frac{13}{-1}\right)f=\frac{42}{-1}$$

Whakawehe -13 ki te -1.

$$f^{2}+13f=\frac{42}{-1}$$

Whakawehe 42 ki te -1.

$$f^{2}+13f=-42$$

Whakawehea te 13, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{13}{2}. Nā, tāpiria te pūrua o te \frac{13}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$f^{2}+13f+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Pūruatia \frac{13}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$f^{2}+13f+\frac{169}{4}=-42+\frac{169}{4}$$

Tāpiri -42 ki te \frac{169}{4}.

$$f^{2}+13f+\frac{169}{4}=\frac{1}{4}$$

Tauwehea f^{2}+13f+\frac{169}{4}. 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}.

$$\left(f+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(f+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$f+\frac{13}{2}=\frac{1}{2} f+\frac{13}{2}=-\frac{1}{2}$$

Me tango \frac{13}{2} mai i ngā taha e rua o te whārite.

$$f=-6 f=-7$$