Réitigh (5sqrt{2}-e)(3sqrt{2}+e)=fsqrt{2}-6

$$( 5 \sqrt { 2 } - e ) ( 3 \sqrt { 2 } + e ) = f \sqrt { 2 } - 6$$

Réitigh do f.

$$f=-\frac{\sqrt{2}e^{2}}{2}+2e+18\sqrt{2}\approx 25.667556106$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Cuir an t-airí dáileacháin i bhfeidhm trí gach téarma de 5\sqrt{2}-e a iolrú faoi gach téarma de 3\sqrt{2}+e.

$$15\left(\sqrt{2}\right)^{2}+5\sqrt{2}e-3e\sqrt{2}-e^{2}=f\sqrt{2}-6$$

Is é 2 uimhir chearnach \sqrt{2}.

$$15\times 2+5\sqrt{2}e-3e\sqrt{2}-e^{2}=f\sqrt{2}-6$$

Méadaigh 15 agus 2 chun 30 a fháil.

$$30+5\sqrt{2}e-3e\sqrt{2}-e^{2}=f\sqrt{2}-6$$

Comhcheangail 5\sqrt{2}e agus -3e\sqrt{2} chun 2\sqrt{2}e a fháil.

$$30+2\sqrt{2}e-e^{2}=f\sqrt{2}-6$$

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

$$f\sqrt{2}-6=30+2\sqrt{2}e-e^{2}$$

Cuir 6 leis an dá thaobh.

$$f\sqrt{2}=30+2\sqrt{2}e-e^{2}+6$$

Suimigh 30 agus 6 chun 36 a fháil.

$$f\sqrt{2}=36+2\sqrt{2}e-e^{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\sqrt{2}f=2e\sqrt{2}-e^{2}+36$$

Roinn an dá thaobh faoi \sqrt{2}.

$$\frac{\sqrt{2}f}{\sqrt{2}}=\frac{2e\sqrt{2}-e^{2}+36}{\sqrt{2}}$$

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

$$f=\frac{2e\sqrt{2}-e^{2}+36}{\sqrt{2}}$$

Roinn 36+2e\sqrt{2}-e^{2} faoi \sqrt{2}.

$$f=\frac{\sqrt{2}\left(2e\sqrt{2}-e^{2}+36\right)}{2}$$