(sqrt{10}-sqrt{2})^2 eching

$$( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }$$

Baholash

$$12-4\sqrt{5}\approx 3,05572809$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{10}-\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

\sqrt{10} kvadrati – 10.

$$10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

Faktor: 10=2\times 5. \sqrt{2\times 5} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{5} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

$$10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}$$

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

$$10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}$$

\sqrt{2} kvadrati – 2.

$$10-4\sqrt{5}+2$$

12 olish uchun 10 va 2'ni qo'shing.

$$12-4\sqrt{5}$$

Kengaytirish

$$12-4\sqrt{5}$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{10}-\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

\sqrt{10} kvadrati – 10.

$$10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

Faktor: 10=2\times 5. \sqrt{2\times 5} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{5} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

$$10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}$$

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

$$10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}$$

\sqrt{2} kvadrati – 2.

$$10-4\sqrt{5}+2$$

12 olish uchun 10 va 2'ni qo'shing.

$$12-4\sqrt{5}$$