frac{sqrt{-2}+1}{sqrt{-2}-1} eching

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

Baholash

$$\text{Indeterminate}$$

Yechim qadamlari

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

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

Hisoblang: \left(\sqrt{-2}-1\right)\left(\sqrt{-2}+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

\sqrt{-2} kvadratini chiqarish. 1 kvadratini chiqarish.

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

-3 olish uchun -2 dan 1 ni ayirish.

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

\left(\sqrt{-2}+1\right)^{2} hosil qilish uchun \sqrt{-2}+1 va \sqrt{-2}+1 ni ko'paytirish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{-2}+1\right)^{2} kengaytirilishi uchun ishlating.

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

2 daraja ko‘rsatkichini \sqrt{-2} ga hisoblang va -2 ni qiymatni oling.

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

-1 olish uchun -2 va 1'ni qo'shing.

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

Surat va maxrajini -1 ga ko‘paytiring.

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

Baholash (complex solution)

$$\frac{-2\sqrt{2}i+1}{3}\approx 0,333333333-0,942809042i$$

Ashyoviy qism (complex solution)

$$\frac{1}{3} = 0,3333333333333333$$