Whakaoti i te 4x-3/2x+1-102x-1/4x-3=3

$$\frac{ 4x-3 }{ 2x+1 } -10 \frac{ 2x-1 }{ 4x-3 } =3$$

Whakaoti mō x

$$x=\frac{5\sqrt{57}}{48}-\frac{3}{16}\approx 0.598941087$$

$$x=-\frac{5\sqrt{57}}{48}-\frac{3}{16}\approx -0.973941087$$

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},\frac{3}{4} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(4x-3\right)\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+1,4x-3.

$$\left(4x-3\right)\left(4x-3\right)-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakareatia te 4x-3 ki te 4x-3, ka \left(4x-3\right)^{2}.

$$\left(4x-3\right)^{2}-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x-3\right)^{2}.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 4x-3.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=\left(12x-9\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 12x-9 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=24x^{2}-6x-9$$

Tangohia te 24x^{2} mai i ngā taha e rua.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)-24x^{2}=-6x-9$$

Me tāpiri te 6x ki ngā taha e rua.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)-24x^{2}+6x=-9$$

Me tāpiri te 9 ki ngā taha e rua.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)-24x^{2}+6x+9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -10 ki te 2x+1.

$$16x^{2}-24x+9+\left(-20x-10\right)\left(2x-1\right)-24x^{2}+6x+9=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -20x-10 ki te 2x-1 ka whakakotahi i ngā kupu rite.

$$16x^{2}-24x+9-40x^{2}+10-24x^{2}+6x+9=0$$

Pahekotia te 16x^{2} me -40x^{2}, ka -24x^{2}.

$$-24x^{2}-24x+9+10-24x^{2}+6x+9=0$$

Tāpirihia te 9 ki te 10, ka 19.

$$-24x^{2}-24x+19-24x^{2}+6x+9=0$$

Pahekotia te -24x^{2} me -24x^{2}, ka -48x^{2}.

$$-48x^{2}-24x+19+6x+9=0$$

Pahekotia te -24x me 6x, ka -18x.

$$-48x^{2}-18x+19+9=0$$

Tāpirihia te 19 ki te 9, ka 28.

$$-48x^{2}-18x+28=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -48 mō a, -18 mō b, me 28 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-48\right)\times 28}}{2\left(-48\right)}$$

Pūrua -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\left(-48\right)\times 28}}{2\left(-48\right)}$$

Whakareatia -4 ki te -48.

$$x=\frac{-\left(-18\right)±\sqrt{324+192\times 28}}{2\left(-48\right)}$$

Whakareatia 192 ki te 28.

$$x=\frac{-\left(-18\right)±\sqrt{324+5376}}{2\left(-48\right)}$$

Tāpiri 324 ki te 5376.

$$x=\frac{-\left(-18\right)±\sqrt{5700}}{2\left(-48\right)}$$

Tuhia te pūtakerua o te 5700.

$$x=\frac{-\left(-18\right)±10\sqrt{57}}{2\left(-48\right)}$$

Ko te tauaro o -18 ko 18.

$$x=\frac{18±10\sqrt{57}}{2\left(-48\right)}$$

Whakareatia 2 ki te -48.

$$x=\frac{18±10\sqrt{57}}{-96}$$

Nā, me whakaoti te whārite x=\frac{18±10\sqrt{57}}{-96} ina he tāpiri te ±. Tāpiri 18 ki te 10\sqrt{57}.

$$x=\frac{10\sqrt{57}+18}{-96}$$

Whakawehe 18+10\sqrt{57} ki te -96.

$$x=-\frac{5\sqrt{57}}{48}-\frac{3}{16}$$

Nā, me whakaoti te whārite x=\frac{18±10\sqrt{57}}{-96} ina he tango te ±. Tango 10\sqrt{57} mai i 18.

$$x=\frac{18-10\sqrt{57}}{-96}$$

Whakawehe 18-10\sqrt{57} ki te -96.

$$x=\frac{5\sqrt{57}}{48}-\frac{3}{16}$$

Kua oti te whārite te whakatau.

$$x=-\frac{5\sqrt{57}}{48}-\frac{3}{16} x=\frac{5\sqrt{57}}{48}-\frac{3}{16}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},\frac{3}{4} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(4x-3\right)\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+1,4x-3.

$$\left(4x-3\right)\left(4x-3\right)-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakareatia te 4x-3 ki te 4x-3, ka \left(4x-3\right)^{2}.

$$\left(4x-3\right)^{2}-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x-3\right)^{2}.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=3\left(4x-3\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 4x-3.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=\left(12x-9\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 12x-9 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)=24x^{2}-6x-9$$

Tangohia te 24x^{2} mai i ngā taha e rua.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)-24x^{2}=-6x-9$$

Me tāpiri te 6x ki ngā taha e rua.

$$16x^{2}-24x+9-10\left(2x+1\right)\left(2x-1\right)-24x^{2}+6x=-9$$

Whakamahia te āhuatanga tohatoha hei whakarea te -10 ki te 2x+1.

$$16x^{2}-24x+9+\left(-20x-10\right)\left(2x-1\right)-24x^{2}+6x=-9$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -20x-10 ki te 2x-1 ka whakakotahi i ngā kupu rite.

$$16x^{2}-24x+9-40x^{2}+10-24x^{2}+6x=-9$$

Pahekotia te 16x^{2} me -40x^{2}, ka -24x^{2}.

$$-24x^{2}-24x+9+10-24x^{2}+6x=-9$$

Tāpirihia te 9 ki te 10, ka 19.

$$-24x^{2}-24x+19-24x^{2}+6x=-9$$

Pahekotia te -24x^{2} me -24x^{2}, ka -48x^{2}.

$$-48x^{2}-24x+19+6x=-9$$

Pahekotia te -24x me 6x, ka -18x.

$$-48x^{2}-18x+19=-9$$

Tangohia te 19 mai i ngā taha e rua.

$$-48x^{2}-18x=-9-19$$

Tangohia te 19 i te -9, ka -28.

$$-48x^{2}-18x=-28$$

Whakawehea ngā taha e rua ki te -48.

$$\frac{-48x^{2}-18x}{-48}=-\frac{28}{-48}$$

Mā te whakawehe ki te -48 ka wetekia te whakareanga ki te -48.

$$x^{2}+\left(-\frac{18}{-48}\right)x=-\frac{28}{-48}$$

Whakahekea te hautanga \frac{-18}{-48} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x^{2}+\frac{3}{8}x=-\frac{28}{-48}$$

Whakahekea te hautanga \frac{-28}{-48} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x^{2}+\frac{3}{8}x=\frac{7}{12}$$

Whakawehea te \frac{3}{8}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{16}. Nā, tāpiria te pūrua o te \frac{3}{16} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{3}{8}x+\left(\frac{3}{16}\right)^{2}=\frac{7}{12}+\left(\frac{3}{16}\right)^{2}$$

Pūruatia \frac{3}{16} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{7}{12}+\frac{9}{256}$$

Tāpiri \frac{7}{12} ki te \frac{9}{256} 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.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{475}{768}$$

Tauwehea te x^{2}+\frac{3}{8}x+\frac{9}{256}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{16}\right)^{2}=\frac{475}{768}$$

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

$$\sqrt{\left(x+\frac{3}{16}\right)^{2}}=\sqrt{\frac{475}{768}}$$

Whakarūnātia.

$$x+\frac{3}{16}=\frac{5\sqrt{57}}{48} x+\frac{3}{16}=-\frac{5\sqrt{57}}{48}$$

Me tango \frac{3}{16} mai i ngā taha e rua o te whārite.

$$x=\frac{5\sqrt{57}}{48}-\frac{3}{16} x=-\frac{5\sqrt{57}}{48}-\frac{3}{16}$$