Whakaoti i te sqrt{x}+sqrt{x+1}=3

$$\sqrt { x } + \sqrt { x + 1 } = 3$$

Whakaoti mō x

$$x = \frac{16}{9} = 1\frac{7}{9} \approx 1.777777778$$

Ngā Upane Otinga

Me tango \sqrt{x+1} mai i ngā taha e rua o te whārite.

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

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

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

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

Tāpirihia te 9 ki te 1, ka 10.

$$x=10-6\sqrt{x+1}+x$$

Me tāpiri te 6\sqrt{x+1} ki ngā taha e rua.

$$x+6\sqrt{x+1}=10+x$$

Tangohia te x mai i ngā taha e rua.

$$x+6\sqrt{x+1}-x=10$$

Pahekotia te x me -x, ka 0.

$$6\sqrt{x+1}=10$$

Whakawehea ngā taha e rua ki te 6.

$$\sqrt{x+1}=\frac{10}{6}$$

Whakahekea te hautanga \frac{10}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$\sqrt{x+1}=\frac{5}{3}$$

Pūruatia ngā taha e rua o te whārite.

$$x+1=\frac{25}{9}$$

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

$$x+1-1=\frac{25}{9}-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$x=\frac{25}{9}-1$$

Tango 1 mai i \frac{25}{9}.

$$x=\frac{16}{9}$$

Whakakapia te \frac{16}{9} mō te x i te whārite \sqrt{x}+\sqrt{x+1}=3.

$$\sqrt{\frac{16}{9}}+\sqrt{\frac{16}{9}+1}=3$$

Whakarūnātia. Ko te uara x=\frac{16}{9} kua ngata te whārite.

$$3=3$$

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

$$x=\frac{16}{9}$$