12i/1+i eching

$$\frac { 12 i } { 1 + i }$$

Baholash

$$6+6i$$

Yechim qadamlari

Ham hisoblagich, ham maxrajni maxraj kompleksiga murakkablash orqali ko'paytirish, 1-i.

$$\frac{12i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

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

$$\frac{12i\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$\frac{12i\left(1-i\right)}{2}$$

12i ni 1-i marotabaga ko'paytirish.

$$\frac{12i\times 1+12\left(-1\right)i^{2}}{2}$$

Ta’rifi bo‘yicha, i^{2} – bu -1.

$$\frac{12i\times 1+12\left(-1\right)\left(-1\right)}{2}$$

12i\times 1+12\left(-1\right)\left(-1\right) ichidagi ko‘paytirishlarni bajaring. Shartlarni qayta saralash.

$$\frac{12+12i}{2}$$

6+6i ni olish uchun 12+12i ni 2 ga bo‘ling.

$$6+6i$$

Ashyoviy qism

$$6$$

Yechim qadamlari

\frac{12i}{1+i}ning surat va maxrajini murakkab tutash maxraj 1-i bilan ko‘paytiring.

$$Re(\frac{12i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

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

$$Re(\frac{12i\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{12i\left(1-i\right)}{2})$$

12i ni 1-i marotabaga ko'paytirish.

$$Re(\frac{12i\times 1+12\left(-1\right)i^{2}}{2})$$

Ta’rifi bo‘yicha, i^{2} – bu -1.

$$Re(\frac{12i\times 1+12\left(-1\right)\left(-1\right)}{2})$$

12i\times 1+12\left(-1\right)\left(-1\right) ichidagi ko‘paytirishlarni bajaring. Shartlarni qayta saralash.

$$Re(\frac{12+12i}{2})$$

6+6i ni olish uchun 12+12i ni 2 ga bo‘ling.

$$Re(6+6i)$$

6+6i ning real qismi – 6.

$$6$$