{left(1/3-sqrt{2}right)}^2 eching

$${ \left( \frac{ 1 }{ 3- \sqrt{ 2 } } \right) }^{ 2 }$$

Baholash

$$\frac{6\sqrt{2}+11}{49}\approx 0,397658804$$

Yechim qadamlari

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

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

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

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

3 kvadratini chiqarish. \sqrt{2} kvadratini chiqarish.

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

7 olish uchun 9 dan 2 ni ayirish.

$$\left(\frac{3+\sqrt{2}}{7}\right)^{2}$$

\frac{3+\sqrt{2}}{7}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7^{2}}$$

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

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7^{2}}$$

\sqrt{2} kvadrati – 2.

$$\frac{9+6\sqrt{2}+2}{7^{2}}$$

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

$$\frac{11+6\sqrt{2}}{7^{2}}$$

2 daraja ko‘rsatkichini 7 ga hisoblang va 49 ni qiymatni oling.

$$\frac{11+6\sqrt{2}}{49}$$