Réitigh frac{10-sqrt{18}}{sqrt{2}}=a+bsqrt{2}

$$\frac { 10 - \sqrt { 18 } } { \sqrt { 2 } } = a + b \sqrt { 2 }$$

Réitigh do b.

$$b=-\frac{\sqrt{2}\left(a+3-5\sqrt{2}\right)}{2}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Fachtóirigh 18=3^{2}\times 2. Athscríobh fréamh cearnach an toraidh \sqrt{3^{2}\times 2} mar thoradh na bhfréamhacha cearnacha \sqrt{3^{2}}\sqrt{2}. Tóg fréamh chearnach 3^{2}.

$$\frac{10-3\sqrt{2}}{\sqrt{2}}=a+b\sqrt{2}$$

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{2} chun ainmneoir \frac{10-3\sqrt{2}}{\sqrt{2}} a thiontú in uimhir chóimheasta.

$$\frac{\left(10-3\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}=a+b\sqrt{2}$$

Is é 2 uimhir chearnach \sqrt{2}.

$$\frac{\left(10-3\sqrt{2}\right)\sqrt{2}}{2}=a+b\sqrt{2}$$

Úsáid an t-airí dáileach chun 10-3\sqrt{2} a mhéadú faoi \sqrt{2}.

$$\frac{10\sqrt{2}-3\left(\sqrt{2}\right)^{2}}{2}=a+b\sqrt{2}$$

Is é 2 uimhir chearnach \sqrt{2}.

$$\frac{10\sqrt{2}-3\times 2}{2}=a+b\sqrt{2}$$

Méadaigh -3 agus 2 chun -6 a fháil.

$$\frac{10\sqrt{2}-6}{2}=a+b\sqrt{2}$$

Roinn 10\sqrt{2}-6 faoi 2 chun 5\sqrt{2}-3 a fháil.

$$5\sqrt{2}-3=a+b\sqrt{2}$$

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

$$a+b\sqrt{2}=5\sqrt{2}-3$$

Bain a ón dá thaobh.

$$b\sqrt{2}=5\sqrt{2}-3-a$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\sqrt{2}b=-a+5\sqrt{2}-3$$

Roinn an dá thaobh faoi \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+5\sqrt{2}-3}{\sqrt{2}}$$

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

$$b=\frac{-a+5\sqrt{2}-3}{\sqrt{2}}$$

Roinn 5\sqrt{2}-a-3 faoi \sqrt{2}.

$$b=\frac{\sqrt{2}\left(-a+5\sqrt{2}-3\right)}{2}$$

Réitigh do a.

$$a=-\sqrt{2}b+5\sqrt{2}-3$$