Whakaoti i te frac{10-sqrt{18}}{sqrt{2}}=a+bsqrt{2}

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

Whakaoti mō b

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Tauwehea te 18=3^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{3^{2}\times 2} hei hua o ngā pūtake rua \sqrt{3^{2}}\sqrt{2}. Tuhia te pūtakerua o te 3^{2}.

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

Whakangāwaritia te tauraro o \frac{10-3\sqrt{2}}{\sqrt{2}} mā te whakarea i te taurunga me te tauraro ki te \sqrt{2}.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 10-3\sqrt{2} ki te \sqrt{2}.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te -3 ki te 2, ka -6.

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

Whakawehea ia wā o 10\sqrt{2}-6 ki te 2, kia riro ko 5\sqrt{2}-3.

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

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

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

Tangohia te a mai i ngā taha e rua.

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

He hanga arowhānui tō te whārite.

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

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

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

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

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

Whakawehe 5\sqrt{2}-a-3 ki te \sqrt{2}.

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

Whakaoti mō a

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