Réitigh (1+sqrt{2})^4=a+bsqrt{2}

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

Réitigh do a.

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

Céimeanna Réitigh

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain b\sqrt{2} ón dá thaobh.

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

Athordaigh na téarmaí.

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

Réitigh do b.

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain a ón dá thaobh.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi \sqrt{2}.

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

Má roinntear é faoi \sqrt{2} cuirtear an iolrúchán faoi \sqrt{2} ar ceal.

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

Roinn 17+12\sqrt{2}-a faoi \sqrt{2}.

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