1/2-i eching

$$\frac { 1 } { 2 - i }$$

Baholash

$$\frac{2}{5}+\frac{1}{5}i=0,4+0,2i$$

Yechim qadamlari

Ham hisoblagich, ham maxrajni maxraj kompleksiga murakkablash orqali ko'paytirish, 2+i.

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

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

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

Ta’rifi bo‘yicha, i^{2} – bu -1. Maxrajini hisoblang.

$$\frac{1\left(2+i\right)}{5}$$

2+i hosil qilish uchun 1 va 2+i ni ko'paytirish.

$$\frac{2+i}{5}$$

\frac{2}{5}+\frac{1}{5}i ni olish uchun 2+i ni 5 ga bo‘ling.

$$\frac{2}{5}+\frac{1}{5}i$$

Ashyoviy qism

$$\frac{2}{5} = 0,4$$

Yechim qadamlari

\frac{1}{2-i}ning surat va maxrajini murakkab tutash maxraj 2+i bilan ko‘paytiring.

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

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

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

Ta’rifi bo‘yicha, i^{2} – bu -1. Maxrajini hisoblang.

$$Re(\frac{1\left(2+i\right)}{5})$$

2+i hosil qilish uchun 1 va 2+i ni ko'paytirish.

$$Re(\frac{2+i}{5})$$

\frac{2}{5}+\frac{1}{5}i ni olish uchun 2+i ni 5 ga bo‘ling.

$$Re(\frac{2}{5}+\frac{1}{5}i)$$

\frac{2}{5}+\frac{1}{5}i ning real qismi – \frac{2}{5}.

$$\frac{2}{5}$$