2019=x(y^2+y+1) eching

$$2019 = x ( y ^ { 2 } + y + 1 )$$

x uchun yechish (complex solution)

$$x=\frac{2019}{y^{2}+y+1}$$

$$y\neq \frac{-1+\sqrt{3}i}{2}\text{ and }y\neq \frac{-\sqrt{3}i-1}{2}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

x ga y^{2}+y+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2019=xy^{2}+xy+x$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$xy^{2}+xy+x=2019$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(y^{2}+y+1\right)x=2019$$

Ikki tarafini y^{2}+y+1 ga bo‘ling.

$$\frac{\left(y^{2}+y+1\right)x}{y^{2}+y+1}=\frac{2019}{y^{2}+y+1}$$

y^{2}+y+1 ga bo'lish y^{2}+y+1 ga ko'paytirishni bekor qiladi.

$$x=\frac{2019}{y^{2}+y+1}$$

x uchun yechish

$$x=\frac{2019}{y^{2}+y+1}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

x ga y^{2}+y+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2019=xy^{2}+xy+x$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$xy^{2}+xy+x=2019$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(y^{2}+y+1\right)x=2019$$

Ikki tarafini y^{2}+y+1 ga bo‘ling.

$$\frac{\left(y^{2}+y+1\right)x}{y^{2}+y+1}=\frac{2019}{y^{2}+y+1}$$

y^{2}+y+1 ga bo'lish y^{2}+y+1 ga ko'paytirishni bekor qiladi.

$$x=\frac{2019}{y^{2}+y+1}$$

y uchun yechish (complex solution)

$$y=\frac{\sqrt{8076x-3x^{2}}}{2x}-\frac{1}{2}$$

$$y=-\frac{\sqrt{8076x-3x^{2}}}{2x}-\frac{1}{2}\text{, }x\neq 0$$

y uchun yechish

$$y=\frac{\sqrt{-3+\frac{8076}{x}}-1}{2}$$

$$y=\frac{-\sqrt{-3+\frac{8076}{x}}-1}{2}\text{, }x>0\text{ and }x\leq 2692$$