12=(1-3x)(1-3x)d+(1+3x)(1+3x) eching

$$12 = ( 1 - 3 x ) ( 1 - 3 x ) d + ( 1 + 3 x ) ( 1 + 3 x )$$

d uchun yechish

$$d=-\frac{9x^{2}+6x-11}{\left(1-3x\right)^{2}}$$

$$x\neq \frac{1}{3}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(1-3x\right)^{2} hosil qilish uchun 1-3x va 1-3x ni ko'paytirish.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)\left(1+3x\right)$$

\left(1+3x\right)^{2} hosil qilish uchun 1+3x va 1+3x ni ko'paytirish.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(1-3x\right)^{2} kengaytirilishi uchun ishlating.

$$12=\left(1-6x+9x^{2}\right)d+\left(1+3x\right)^{2}$$

1-6x+9x^{2} ga d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12=d-6xd+9x^{2}d+\left(1+3x\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+3x\right)^{2} kengaytirilishi uchun ishlating.

$$12=d-6xd+9x^{2}d+1+6x+9x^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$d-6xd+9x^{2}d+1+6x+9x^{2}=12$$

Ikkala tarafdan 1 ni ayirish.

$$d-6xd+9x^{2}d+6x+9x^{2}=12-1$$

11 olish uchun 12 dan 1 ni ayirish.

$$d-6xd+9x^{2}d+6x+9x^{2}=11$$

Ikkala tarafdan 6x ni ayirish.

$$d-6xd+9x^{2}d+9x^{2}=11-6x$$

Ikkala tarafdan 9x^{2} ni ayirish.

$$d-6xd+9x^{2}d=11-6x-9x^{2}$$

d'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1-6x+9x^{2}\right)d=11-6x-9x^{2}$$

Tenglama standart shaklda.

$$\left(9x^{2}-6x+1\right)d=11-6x-9x^{2}$$

Ikki tarafini 1-6x+9x^{2} ga bo‘ling.

$$\frac{\left(9x^{2}-6x+1\right)d}{9x^{2}-6x+1}=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

1-6x+9x^{2} ga bo'lish 1-6x+9x^{2} ga ko'paytirishni bekor qiladi.

$$d=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

11-6x-9x^{2} ni 1-6x+9x^{2} ga bo'lish.

$$d=\frac{11-6x-9x^{2}}{\left(3x-1\right)^{2}}$$

x uchun yechish (complex solution)

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\\x=1\text{, }&d=-1\end{matrix}\right,$$

x uchun yechish

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\text{ and }d\geq -\frac{3}{2}\\x=1\text{, }&d=-1\end{matrix}\right,$$