Datrys {left(1/3-sqrt{2}right)}^2

$${ \left( \frac{ 1 }{ 3- \sqrt{ 2 } } \right) }^{ 2 }$$

Enrhifo

$$\frac{6\sqrt{2}+11}{49}\approx 0.397658804$$

Camau Ateb

Mae'n rhesymoli enwadur \frac{1}{3-\sqrt{2}} drwy luosi'r rhifiadur a'r enwadur â 3+\sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\right)^{2}$$

Ystyriwch \left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right). Gellir trawsnewid lluosi yn wahaniaeth rhwng sgwariau drwy ddefnyddio’r rheol: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{3+\sqrt{2}}{3^{2}-\left(\sqrt{2}\right)^{2}}\right)^{2}$$

Sgwâr 3. Sgwâr \sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{9-2}\right)^{2}$$

Tynnu 2 o 9 i gael 7.

$$\left(\frac{3+\sqrt{2}}{7}\right)^{2}$$

I godi \frac{3+\sqrt{2}}{7} i bŵer, codwch y rhifiadur a'r enwadur i bŵer ac yna rhannwch nhw.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7^{2}}$$

Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(3+\sqrt{2}\right)^{2}.

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7^{2}}$$

Sgwâr \sqrt{2} yw 2.

$$\frac{9+6\sqrt{2}+2}{7^{2}}$$

Adio 9 a 2 i gael 11.

$$\frac{11+6\sqrt{2}}{7^{2}}$$

Cyfrifo 7 i bŵer 2 a chael 49.

$$\frac{11+6\sqrt{2}}{49}$$