(sqrt{2}-3)^2+6*(sqrt{2}-3)+5 eching

$$( \sqrt { 2 } - 3 ) ^ { 2 } + 6 \times ( \sqrt { 2 } - 3 ) + 5$$

Baholash

$$-2$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{2}-3\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{2}\right)^{2}-6\sqrt{2}+9+6\left(\sqrt{2}-3\right)+5$$

\sqrt{2} kvadrati – 2.

$$2-6\sqrt{2}+9+6\left(\sqrt{2}-3\right)+5$$

11 olish uchun 2 va 9'ni qo'shing.

$$11-6\sqrt{2}+6\left(\sqrt{2}-3\right)+5$$

6 ga \sqrt{2}-3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$11-6\sqrt{2}+6\sqrt{2}-18+5$$

0 ni olish uchun -6\sqrt{2} va 6\sqrt{2} ni birlashtirish.

$$11-18+5$$

-7 olish uchun 11 dan 18 ni ayirish.

$$-7+5$$

-2 olish uchun -7 va 5'ni qo'shing.

$$-2$$

Omil

$$-2$$