ax^2+b^2-2ab^2=(b-ab)x eching

$$a x ^ { 2 } + b ^ { 2 } - 2 a b ^ { 2 } = ( b - a b ) x$$

a uchun yechish (complex solution)

$$\left\{\begin{matrix}a=\frac{b}{x+2b}\text{, }&x\neq -2b\\a\in \mathrm{C}\text{, }&b=x\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

b-ab ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$ax^{2}+b^{2}-2ab^{2}=bx-abx$$

abx ni ikki tarafga qo’shing.

$$ax^{2}+b^{2}-2ab^{2}+abx=bx$$

Ikkala tarafdan b^{2} ni ayirish.

$$ax^{2}-2ab^{2}+abx=bx-b^{2}$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x^{2}-2b^{2}+bx\right)a=bx-b^{2}$$

Tenglama standart shaklda.

$$\left(x^{2}+bx-2b^{2}\right)a=bx-b^{2}$$

Ikki tarafini x^{2}-2b^{2}+bx ga bo‘ling.

$$\frac{\left(x^{2}+bx-2b^{2}\right)a}{x^{2}+bx-2b^{2}}=\frac{b\left(x-b\right)}{x^{2}+bx-2b^{2}}$$

x^{2}-2b^{2}+bx ga bo'lish x^{2}-2b^{2}+bx ga ko'paytirishni bekor qiladi.

$$a=\frac{b\left(x-b\right)}{x^{2}+bx-2b^{2}}$$

b\left(x-b\right) ni x^{2}-2b^{2}+bx ga bo'lish.

$$a=\frac{b}{x+2b}$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{b}{x+2b}\text{, }&x\neq -2b\\a\in \mathrm{R}\text{, }&x=b\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

b-ab ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$ax^{2}+b^{2}-2ab^{2}=bx-abx$$

abx ni ikki tarafga qo’shing.

$$ax^{2}+b^{2}-2ab^{2}+abx=bx$$

Ikkala tarafdan b^{2} ni ayirish.

$$ax^{2}-2ab^{2}+abx=bx-b^{2}$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x^{2}-2b^{2}+bx\right)a=bx-b^{2}$$

Tenglama standart shaklda.

$$\left(x^{2}+bx-2b^{2}\right)a=bx-b^{2}$$

Ikki tarafini x^{2}-2b^{2}+bx ga bo‘ling.

$$\frac{\left(x^{2}+bx-2b^{2}\right)a}{x^{2}+bx-2b^{2}}=\frac{b\left(x-b\right)}{x^{2}+bx-2b^{2}}$$

x^{2}-2b^{2}+bx ga bo'lish x^{2}-2b^{2}+bx ga ko'paytirishni bekor qiladi.

$$a=\frac{b\left(x-b\right)}{x^{2}+bx-2b^{2}}$$

b\left(x-b\right) ni x^{2}-2b^{2}+bx ga bo'lish.

$$a=\frac{b}{x+2b}$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}\\b=x\text{, }&\text{unconditionally}\\b=\frac{ax}{1-2a}\text{, }&a\neq \frac{1}{2}\\b\in \mathrm{C}\text{, }&x=0\text{ and }a=\frac{1}{2}\end{matrix}\right,$$

b uchun yechish

$$\left\{\begin{matrix}\\b=x\text{, }&\text{unconditionally}\\b=\frac{ax}{1-2a}\text{, }&a\neq \frac{1}{2}\\b\in \mathrm{R}\text{, }&x=0\text{ and }a=\frac{1}{2}\end{matrix}\right,$$