Whakaoti i te sqrt{2x+1}+1=x

$$\sqrt{ 2x+1 } +1=x$$

Whakaoti mō x

$$x=4$$

Ngā Upane Otinga

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

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

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

$$2x+1=x^{2}-2x+1$$

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

$$2x+1-x^{2}=-2x+1$$

Me tāpiri te 2x ki ngā taha e rua.

$$2x+1-x^{2}+2x=1$$

Pahekotia te 2x me 2x, ka 4x.

$$4x+1-x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$4x+1-x^{2}-1=0$$

Tangohia te 1 i te 1, ka 0.

$$4x-x^{2}=0$$

Tauwehea te x.

$$x\left(4-x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 4-x=0.

$$x=0 x=4$$

Whakakapia te 0 mō te x i te whārite \sqrt{2x+1}+1=x.

$$\sqrt{2\times 0+1}+1=0$$

Whakarūnātia. Ko te uara x=0 kāore e ngata ana ki te whārite.

$$2=0$$

Whakakapia te 4 mō te x i te whārite \sqrt{2x+1}+1=x.

$$\sqrt{2\times 4+1}+1=4$$

Whakarūnātia. Ko te uara x=4 kua ngata te whārite.

$$4=4$$

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

$$x=4$$