1/3-sqrt{8}-2sqrt{2}+6 eching

$$\frac { 1 } { 3 - \sqrt { 8 } } - 2 \sqrt { 2 } + 6$$

Baholash

$$9$$

Yechim qadamlari

Faktor: 8=2^{2}\times 2. \sqrt{2^{2}\times 2} koʻpaytmasining kvadrat ildizini \sqrt{2^{2}}\sqrt{2} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing. 2^{2} ning kvadrat ildizini chiqarish.

$$\frac{1}{3-2\sqrt{2}}-2\sqrt{2}+6$$

\frac{1}{3-2\sqrt{2}} maxrajini 3+2\sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

Hisoblang: \left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

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

\left(-2\sqrt{2}\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini -2 ga hisoblang va 4 ni qiymatni oling.

$$\frac{3+2\sqrt{2}}{9-4\left(\sqrt{2}\right)^{2}}-2\sqrt{2}+6$$

\sqrt{2} kvadrati – 2.

$$\frac{3+2\sqrt{2}}{9-4\times 2}-2\sqrt{2}+6$$

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

$$\frac{3+2\sqrt{2}}{9-8}-2\sqrt{2}+6$$

1 olish uchun 9 dan 8 ni ayirish.

$$\frac{3+2\sqrt{2}}{1}-2\sqrt{2}+6$$

Har qanday son birga bo‘linganda, natija o‘zi chiqadi.

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

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

$$3+6$$

9 olish uchun 3 va 6'ni qo'shing.

$$9$$

Omil

$$3^{2}$$