left(x+8right)^2=(x+2)y eching

$${ \left(x+8 \right) }^{ 2 } =(x+2)y$$

y uchun yechish

$$y=\frac{\left(x+8\right)^{2}}{x+2}$$

$$x\neq -2$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+8\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+16x+64=\left(x+2\right)y$$

x+2 ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{2}+16x+64=xy+2y$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$xy+2y=x^{2}+16x+64$$

y'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x+2\right)y=x^{2}+16x+64$$

Ikki tarafini x+2 ga bo‘ling.

$$\frac{\left(x+2\right)y}{x+2}=\frac{\left(x+8\right)^{2}}{x+2}$$

x+2 ga bo'lish x+2 ga ko'paytirishni bekor qiladi.

$$y=\frac{\left(x+8\right)^{2}}{x+2}$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{y\left(y-24\right)}+y-16}{2}$$

$$x=\frac{-\sqrt{y\left(y-24\right)}+y-16}{2}$$

x uchun yechish

$$x=\frac{\sqrt{y\left(y-24\right)}+y-16}{2}$$

$$x=\frac{-\sqrt{y\left(y-24\right)}+y-16}{2}\text{, }y\geq 24\text{ or }y\leq 0$$