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

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

Whakaoti mō x

$$x=1$$

Ngā Upane Otinga

Me tango -2 mai i ngā taha e rua o te whārite.

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

Whakareatia ngā taha e rua o te whārite ki te 2.

$$3\sqrt{x}-5+4=2\sqrt{x}$$

Tāpirihia te -5 ki te 4, ka -1.

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

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

Whakarohaina te \left(2\sqrt{x}\right)^{2}.

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

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

$$9x-6\sqrt{x}+1=4\left(\sqrt{x}\right)^{2}$$

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

$$9x-6\sqrt{x}+1=4x$$

Me tango 9x+1 mai i ngā taha e rua o te whārite.

$$-6\sqrt{x}=4x-\left(9x+1\right)$$

Hei kimi i te tauaro o 9x+1, kimihia te tauaro o ia taurangi.

$$-6\sqrt{x}=4x-9x-1$$

Pahekotia te 4x me -9x, ka -5x.

$$-6\sqrt{x}=-5x-1$$

Pūruatia ngā taha e rua o te whārite.

$$\left(-6\sqrt{x}\right)^{2}=\left(-5x-1\right)^{2}$$

Whakarohaina te \left(-6\sqrt{x}\right)^{2}.

$$\left(-6\right)^{2}\left(\sqrt{x}\right)^{2}=\left(-5x-1\right)^{2}$$

Tātaihia te -6 mā te pū o 2, kia riro ko 36.

$$36\left(\sqrt{x}\right)^{2}=\left(-5x-1\right)^{2}$$

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

$$36x=\left(-5x-1\right)^{2}$$

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

$$36x=25x^{2}+10x+1$$

Tangohia te 25x^{2} mai i ngā taha e rua.

$$36x-25x^{2}=10x+1$$

Tangohia te 10x mai i ngā taha e rua.

$$36x-25x^{2}-10x=1$$

Pahekotia te 36x me -10x, ka 26x.

$$26x-25x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$26x-25x^{2}-1=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-25x^{2}+26x-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -25x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=26 ab=-25\left(-1\right)=25$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 25.

$$1,25 5,5$$

Tātaihia te tapeke mō ia takirua.

$$1+25=26 5+5=10$$

Ko te otinga te takirua ka hoatu i te tapeke 26.

$$a=25 b=1$$

Tuhia anō te -25x^{2}+26x-1 hei \left(-25x^{2}+25x\right)+\left(x-1\right).

$$\left(-25x^{2}+25x\right)+\left(x-1\right)$$

Tauwehea te 25x i te tuatahi me te -1 i te rōpū tuarua.

$$25x\left(-x+1\right)-\left(-x+1\right)$$

Whakatauwehea atu te kīanga pātahi -x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+1\right)\left(25x-1\right)$$

Hei kimi otinga whārite, me whakaoti te -x+1=0 me te 25x-1=0.

$$x=1 x=\frac{1}{25}$$

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

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

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$-1=-1$$

Whakakapia te \frac{1}{25} mō te x i te whārite \frac{3\sqrt{x}-5}{2}=\sqrt{x}-2.

$$\frac{3\sqrt{\frac{1}{25}}-5}{2}=\sqrt{\frac{1}{25}}-2$$

Whakarūnātia. Ko te uara x=\frac{1}{25} kāore e ngata ana ki te whārite.

$$-\frac{11}{5}=-\frac{9}{5}$$

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

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

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$-1=-1$$

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

$$x=1$$