-(lambda-1)(lambda^2-blambda+5)=0 eching

$$- ( \lambda - 1 ) ( \lambda ^ { 2 } - b \lambda + 5 ) = 0$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}b=\lambda +\frac{5}{\lambda }\text{, }&\lambda \neq 0\\b\in \mathrm{C}\text{, }&\lambda =1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\lambda -1 teskarisini topish uchun har birining teskarisini toping.

$$\left(-\lambda +1\right)\left(\lambda ^{2}-b\lambda +5\right)=0$$

-\lambda +1 ga \lambda ^{2}-b\lambda +5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-\lambda ^{3}+\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=0$$

\lambda ^{3} ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=\lambda ^{3}$$

5\lambda ni ikki tarafga qo’shing.

$$\lambda ^{2}b+\lambda ^{2}-b\lambda +5=\lambda ^{3}+5\lambda$$

Ikkala tarafdan \lambda ^{2} ni ayirish.

$$\lambda ^{2}b-b\lambda +5=\lambda ^{3}+5\lambda -\lambda ^{2}$$

Ikkala tarafdan 5 ni ayirish.

$$\lambda ^{2}b-b\lambda =\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

b'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

Tenglama standart shaklda.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}-\lambda ^{2}+5\lambda -5$$

Ikki tarafini \lambda ^{2}-\lambda ga bo‘ling.

$$\frac{\left(\lambda ^{2}-\lambda \right)b}{\lambda ^{2}-\lambda }=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

\lambda ^{2}-\lambda ga bo'lish \lambda ^{2}-\lambda ga ko'paytirishni bekor qiladi.

$$b=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

\left(-1+\lambda \right)\left(5+\lambda ^{2}\right) ni \lambda ^{2}-\lambda ga bo'lish.

$$b=\lambda +\frac{5}{\lambda }$$

b uchun yechish

$$\left\{\begin{matrix}b=\lambda +\frac{5}{\lambda }\text{, }&\lambda \neq 0\\b\in \mathrm{R}\text{, }&\lambda =1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\lambda -1 teskarisini topish uchun har birining teskarisini toping.

$$\left(-\lambda +1\right)\left(\lambda ^{2}-b\lambda +5\right)=0$$

-\lambda +1 ga \lambda ^{2}-b\lambda +5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-\lambda ^{3}+\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=0$$

\lambda ^{3} ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$\lambda ^{2}b-5\lambda +\lambda ^{2}-b\lambda +5=\lambda ^{3}$$

5\lambda ni ikki tarafga qo’shing.

$$\lambda ^{2}b+\lambda ^{2}-b\lambda +5=\lambda ^{3}+5\lambda$$

Ikkala tarafdan \lambda ^{2} ni ayirish.

$$\lambda ^{2}b-b\lambda +5=\lambda ^{3}+5\lambda -\lambda ^{2}$$

Ikkala tarafdan 5 ni ayirish.

$$\lambda ^{2}b-b\lambda =\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

b'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}+5\lambda -\lambda ^{2}-5$$

Tenglama standart shaklda.

$$\left(\lambda ^{2}-\lambda \right)b=\lambda ^{3}-\lambda ^{2}+5\lambda -5$$

Ikki tarafini \lambda ^{2}-\lambda ga bo‘ling.

$$\frac{\left(\lambda ^{2}-\lambda \right)b}{\lambda ^{2}-\lambda }=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

\lambda ^{2}-\lambda ga bo'lish \lambda ^{2}-\lambda ga ko'paytirishni bekor qiladi.

$$b=\frac{\left(\lambda -1\right)\left(\lambda ^{2}+5\right)}{\lambda ^{2}-\lambda }$$

\left(-1+\lambda \right)\left(5+\lambda ^{2}\right) ni \lambda ^{2}-\lambda ga bo'lish.

$$b=\lambda +\frac{5}{\lambda }$$

λ uchun yechish (complex solution)

$$\lambda =\frac{-\sqrt{b^{2}-20}+b}{2}$$

$$\lambda =1$$

$$\lambda =\frac{\sqrt{b^{2}-20}+b}{2}$$

λ uchun yechish

$$\left\{\begin{matrix}\\\lambda =1\text{, }&\text{unconditionally}\\\lambda =\frac{\sqrt{b^{2}-20}+b}{2}\text{; }\lambda =\frac{-\sqrt{b^{2}-20}+b}{2}\text{, }&|b|\geq 2\sqrt{5}\end{matrix}\right,$$