Whakaoti i te sqrt{5x-1}-sqrt{3x-2}=sqrt{x-1}

$$\sqrt { 5 x - 1 } - \sqrt { 3 x - 2 } = \sqrt { x - 1 }$$

Whakaoti mō x

$$x=2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(\sqrt{5x-1}-\sqrt{3x-2}\right)^{2}.

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

Tātaihia te \sqrt{5x-1} mā te pū o 2, kia riro ko 5x-1.

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

Tātaihia te \sqrt{3x-2} mā te pū o 2, kia riro ko 3x-2.

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

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

$$8x-1-2\sqrt{5x-1}\sqrt{3x-2}-2=\left(\sqrt{x-1}\right)^{2}$$

Tangohia te 2 i te -1, ka -3.

$$8x-3-2\sqrt{5x-1}\sqrt{3x-2}=\left(\sqrt{x-1}\right)^{2}$$

Tātaihia te \sqrt{x-1} mā te pū o 2, kia riro ko x-1.

$$8x-3-2\sqrt{5x-1}\sqrt{3x-2}=x-1$$

Me tango 8x-3 mai i ngā taha e rua o te whārite.

$$-2\sqrt{5x-1}\sqrt{3x-2}=x-1-\left(8x-3\right)$$

Hei kimi i te tauaro o 8x-3, kimihia te tauaro o ia taurangi.

$$-2\sqrt{5x-1}\sqrt{3x-2}=x-1-8x+3$$

Pahekotia te x me -8x, ka -7x.

$$-2\sqrt{5x-1}\sqrt{3x-2}=-7x-1+3$$

Tāpirihia te -1 ki te 3, ka 2.

$$-2\sqrt{5x-1}\sqrt{3x-2}=-7x+2$$

Pūruatia ngā taha e rua o te whārite.

$$\left(-2\sqrt{5x-1}\sqrt{3x-2}\right)^{2}=\left(-7x+2\right)^{2}$$

Whakarohaina te \left(-2\sqrt{5x-1}\sqrt{3x-2}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{5x-1}\right)^{2}\left(\sqrt{3x-2}\right)^{2}=\left(-7x+2\right)^{2}$$

Tātaihia te -2 mā te pū o 2, kia riro ko 4.

$$4\left(\sqrt{5x-1}\right)^{2}\left(\sqrt{3x-2}\right)^{2}=\left(-7x+2\right)^{2}$$

Tātaihia te \sqrt{5x-1} mā te pū o 2, kia riro ko 5x-1.

$$4\left(5x-1\right)\left(\sqrt{3x-2}\right)^{2}=\left(-7x+2\right)^{2}$$

Tātaihia te \sqrt{3x-2} mā te pū o 2, kia riro ko 3x-2.

$$4\left(5x-1\right)\left(3x-2\right)=\left(-7x+2\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 5x-1.

$$\left(20x-4\right)\left(3x-2\right)=\left(-7x+2\right)^{2}$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 20x-4 ki ia tau o 3x-2.

$$60x^{2}-40x-12x+8=\left(-7x+2\right)^{2}$$

Pahekotia te -40x me -12x, ka -52x.

$$60x^{2}-52x+8=\left(-7x+2\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(-7x+2\right)^{2}.

$$60x^{2}-52x+8=49x^{2}-28x+4$$

Tangohia te 49x^{2} mai i ngā taha e rua.

$$60x^{2}-52x+8-49x^{2}=-28x+4$$

Pahekotia te 60x^{2} me -49x^{2}, ka 11x^{2}.

$$11x^{2}-52x+8=-28x+4$$

Me tāpiri te 28x ki ngā taha e rua.

$$11x^{2}-52x+8+28x=4$$

Pahekotia te -52x me 28x, ka -24x.

$$11x^{2}-24x+8=4$$

Tangohia te 4 mai i ngā taha e rua.

$$11x^{2}-24x+8-4=0$$

Tangohia te 4 i te 8, ka 4.

$$11x^{2}-24x+4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 11x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-24 ab=11\times 4=44$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 44.

$$-1,-44 -2,-22 -4,-11$$

Tātaihia te tapeke mō ia takirua.

$$-1-44=-45 -2-22=-24 -4-11=-15$$

Ko te otinga te takirua ka hoatu i te tapeke -24.

$$a=-22 b=-2$$

Tuhia anō te 11x^{2}-24x+4 hei \left(11x^{2}-22x\right)+\left(-2x+4\right).

$$\left(11x^{2}-22x\right)+\left(-2x+4\right)$$

Tauwehea te 11x i te tuatahi me te -2 i te rōpū tuarua.

$$11x\left(x-2\right)-2\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(11x-2\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te 11x-2=0.

$$x=2 x=\frac{2}{11}$$

Whakakapia te \frac{2}{11} mō te x i te whārite \sqrt{5x-1}-\sqrt{3x-2}=\sqrt{x-1}. Te kīanga \sqrt{5\times \frac{2}{11}-1} kia kore e tautuhitia nā te mea kāore te radicand e noho tōraro.

$$\sqrt{5\times \frac{2}{11}-1}-\sqrt{3\times \frac{2}{11}-2}=\sqrt{\frac{2}{11}-1}$$

Whakakapia te 2 mō te x i te whārite \sqrt{5x-1}-\sqrt{3x-2}=\sqrt{x-1}.

$$\sqrt{5\times 2-1}-\sqrt{3\times 2-2}=\sqrt{2-1}$$

Whakarūnātia. Ko te uara x=2 kua ngata te whārite.

$$1=1$$

Ko te whārite \sqrt{5x-1}-\sqrt{3x-2}=\sqrt{x-1} he rongoā ahurei.

$$x=2$$