Whakaoti i te (1+sqrt{2})^4=a+bsqrt{2}

$$( 1 + \sqrt { 2 } ) ^ { 4 } = a + b \sqrt { 2 }$$

Whakaoti mō a

$$a=\sqrt{2}\left(12-b\right)+17$$

Ngā Upane Otinga

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Tangohia te b\sqrt{2} mai i ngā taha e rua.

$$a=\left(1+\sqrt{2}\right)^{4}-b\sqrt{2}$$

Whakaraupapatia anō ngā kīanga tau.

$$a=-\sqrt{2}b+\left(\sqrt{2}+1\right)^{4}$$

Whakaoti mō b

$$b=-\frac{\sqrt{2}\left(a-12\sqrt{2}-17\right)}{2}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Tangohia te a mai i ngā taha e rua.

$$b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}-a$$

He hanga arowhānui tō te whārite.

$$\sqrt{2}b=-a+\left(\sqrt{2}+1\right)^{4}$$

Whakawehea ngā taha e rua ki te \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

Mā te whakawehe ki te \sqrt{2} ka wetekia te whakareanga ki te \sqrt{2}.

$$b=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

Whakawehe 17+12\sqrt{2}-a ki te \sqrt{2}.

$$b=\frac{\sqrt{2}\left(-a+12\sqrt{2}+17\right)}{2}$$