kz=(2+i)m^2-3(i+1)m-2(1-i) eching

$$k z = ( 2 + i ) m ^ { 2 } - 3 ( i + 1 ) m - 2 ( 1 - i )$$

k uchun yechish

$$\left\{\begin{matrix}k=-\frac{\left(-2-i\right)m^{2}+\left(3+3i\right)m+\left(2-2i\right)}{z}\text{, }&z\neq 0\\k\in \mathrm{C}\text{, }&\left(m=2\text{ or }m=-\frac{1}{5}+\frac{3}{5}i\right)\text{ and }z=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$kz=\left(2+i\right)m^{2}-3\left(i+1\right)m-\left(2-2i\right)$$

-3 ga i+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$kz=\left(2+i\right)m^{2}+\left(-3i-3\right)m-\left(2-2i\right)$$

-3i-3 ga m ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m-\left(2-2i\right)$$

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

$$kz=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

Tenglama standart shaklda.

$$zk=\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)$$

Ikki tarafini z ga bo‘ling.

$$\frac{zk}{z}=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

z ga bo'lish z ga ko'paytirishni bekor qiladi.

$$k=\frac{\left(2+i\right)m^{2}+\left(-3-3i\right)m+\left(-2+2i\right)}{z}$$

m uchun yechish

$$m=\left(\frac{1}{5}-\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$

$$m=\left(-\frac{1}{5}+\frac{1}{10}i\right)\sqrt{\left(8+4i\right)kz+\left(24+10i\right)}+\left(\frac{9}{10}+\frac{3}{10}i\right)$$