ax^2=a+b-bx eching

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

a uchun yechish (complex solution)

$$\left\{\begin{matrix}a=-\frac{b}{x+1}\text{, }&x\neq -1\\a\in \mathrm{C}\text{, }&x=1\text{ or }\left(b=0\text{ and }x=-1\right)\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Ikkala tarafdan a ni ayirish.

$$ax^{2}-a=b-bx$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x^{2}-1\right)a=b-bx$$

Ikki tarafini x^{2}-1 ga bo‘ling.

$$\frac{\left(x^{2}-1\right)a}{x^{2}-1}=\frac{b-bx}{x^{2}-1}$$

x^{2}-1 ga bo'lish x^{2}-1 ga ko'paytirishni bekor qiladi.

$$a=\frac{b-bx}{x^{2}-1}$$

b-bx ni x^{2}-1 ga bo'lish.

$$a=-\frac{b}{x+1}$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}\\b=-a\left(x+1\right)\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&x=1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+b-bx=ax^{2}$$

Ikkala tarafdan a ni ayirish.

$$b-bx=ax^{2}-a$$

b'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1-x\right)b=ax^{2}-a$$

Ikki tarafini 1-x ga bo‘ling.

$$\frac{\left(1-x\right)b}{1-x}=\frac{a\left(x^{2}-1\right)}{1-x}$$

1-x ga bo'lish 1-x ga ko'paytirishni bekor qiladi.

$$b=\frac{a\left(x^{2}-1\right)}{1-x}$$

a\left(x^{2}-1\right) ni 1-x ga bo'lish.

$$b=-a\left(x+1\right)$$

a uchun yechish

$$\left\{\begin{matrix}a=-\frac{b}{x+1}\text{, }&x\neq -1\\a\in \mathrm{R}\text{, }&x=1\text{ or }\left(b=0\text{ and }x=-1\right)\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Ikkala tarafdan a ni ayirish.

$$ax^{2}-a=b-bx$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x^{2}-1\right)a=b-bx$$

Ikki tarafini x^{2}-1 ga bo‘ling.

$$\frac{\left(x^{2}-1\right)a}{x^{2}-1}=\frac{b-bx}{x^{2}-1}$$

x^{2}-1 ga bo'lish x^{2}-1 ga ko'paytirishni bekor qiladi.

$$a=\frac{b-bx}{x^{2}-1}$$

b-bx ni x^{2}-1 ga bo'lish.

$$a=-\frac{b}{x+1}$$

b uchun yechish

$$\left\{\begin{matrix}\\b=-a\left(x+1\right)\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&x=1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+b-bx=ax^{2}$$

Ikkala tarafdan a ni ayirish.

$$b-bx=ax^{2}-a$$

b'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1-x\right)b=ax^{2}-a$$

Ikki tarafini 1-x ga bo‘ling.

$$\frac{\left(1-x\right)b}{1-x}=\frac{a\left(x^{2}-1\right)}{1-x}$$

1-x ga bo'lish 1-x ga ko'paytirishni bekor qiladi.

$$b=\frac{a\left(x^{2}-1\right)}{1-x}$$

a\left(x^{2}-1\right) ni 1-x ga bo'lish.

$$b=-a\left(x+1\right)$$