Issolvi (1+sqrt{2})^4=a+bsqrt{2}

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

Solvi għal a

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

Passi tas-Soluzzjonijiet

Ibdel in-naħat sabiex it-termini varjabbli kollha jkunu fuq in-naħa tax-xellug.

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

Naqqas b\sqrt{2} miż-żewġ naħat.

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

Erġa' ordna t-termini.

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

Solvi għal b

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

Passi biex Issolvi l-Ekwazzjoni Lineari

Ibdel in-naħat sabiex it-termini varjabbli kollha jkunu fuq in-naħa tax-xellug.

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

Naqqas a miż-żewġ naħat.

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

L-ekwazzjoni hija f'forma standard.

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

Iddividi ż-żewġ naħat b'\sqrt{2}.

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

Meta tiddividi b'\sqrt{2} titneħħa l-multiplikazzjoni b'\sqrt{2}.

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

Iddividi 17+12\sqrt{2}-a b'\sqrt{2}.

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