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

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

Baholash

$$\sqrt{2}+8\approx 9,414213562$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{6}+\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{6} kvadrati – 6.

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

Faktor: 6=2\times 3. \sqrt{2\times 3} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{3} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

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

2 hosil qilish uchun \sqrt{2} va \sqrt{2} ni ko'paytirish.

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

4 hosil qilish uchun 2 va 2 ni ko'paytirish.

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

\sqrt{2} kvadrati – 2.

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

8 olish uchun 6 va 2'ni qo'shing.

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

Faktor: 6=2\times 3. \sqrt{2\times 3} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{3} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

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

2 hosil qilish uchun \sqrt{2} va \sqrt{2} ni ko'paytirish.

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

4 hosil qilish uchun 2 va 2 ni ko'paytirish.

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

0 ni olish uchun 4\sqrt{3} va -4\sqrt{3} ni birlashtirish.

$$8+\sqrt{2}$$