Réitigh {left(sqrt{6}+sqrt{2}right)}^2-2sqrt{2}*sqrt{6}+sqrt{2}

$${ \left( \sqrt{ 6 } + \sqrt{ 2 } \right) }^{ 2 } -2 \sqrt{ 2 } \times \sqrt{ 6 } + \sqrt{ 2 }$$

Luacháil

$$\sqrt{2}+8\approx 9.414213562$$

Céimeanna Réitigh

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(\sqrt{6}+\sqrt{2}\right)^{2} a leathnú.

$$\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Is é 6 uimhir chearnach \sqrt{6}.

$$6+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Fachtóirigh 6=2\times 3. Athscríobh fréamh cearnach an toraidh \sqrt{2\times 3} mar thoradh na bhfréamhacha cearnacha \sqrt{2}\sqrt{3}.

$$6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Méadaigh \sqrt{2} agus \sqrt{2} chun 2 a fháil.

$$6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Méadaigh 2 agus 2 chun 4 a fháil.

$$6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Is é 2 uimhir chearnach \sqrt{2}.

$$6+4\sqrt{3}+2-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Suimigh 6 agus 2 chun 8 a fháil.

$$8+4\sqrt{3}-2\sqrt{2}\sqrt{6}+\sqrt{2}$$

Fachtóirigh 6=2\times 3. Athscríobh fréamh cearnach an toraidh \sqrt{2\times 3} mar thoradh na bhfréamhacha cearnacha \sqrt{2}\sqrt{3}.

$$8+4\sqrt{3}-2\sqrt{2}\sqrt{2}\sqrt{3}+\sqrt{2}$$

Méadaigh \sqrt{2} agus \sqrt{2} chun 2 a fháil.

$$8+4\sqrt{3}-2\times 2\sqrt{3}+\sqrt{2}$$

Méadaigh 2 agus 2 chun 4 a fháil.

$$8+4\sqrt{3}-4\sqrt{3}+\sqrt{2}$$

Comhcheangail 4\sqrt{3} agus -4\sqrt{3} chun 0 a fháil.

$$8+\sqrt{2}$$