y=sqrt{x+y} eching

$$y = \sqrt { x + y }$$

x uchun yechish

$$x=y\left(y-1\right)$$

$$y\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{x+y}=y$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$x+y=y^{2}$$

Tenglamaning ikkala tarafidan y ni ayirish.

$$x+y-y=y^{2}-y$$

O‘zidan y ayirilsa 0 qoladi.

$$x=y^{2}-y$$

y^{2} dan y ni ayirish.

$$x=y\left(y-1\right)$$

x uchun yechish (complex solution)

$$x=y\left(y-1\right)$$

$$arg(y)<\pi \text{ or }y=0$$

y uchun yechish (complex solution)

$$\left\{\begin{matrix}y=\frac{-\sqrt{4x+1}+1}{2}\text{, }&arg(\frac{-\sqrt{4x+1}+1}{2})<\pi \text{ or }x=0\\y=\frac{\sqrt{4x+1}+1}{2}\text{, }&arg(\frac{\sqrt{4x+1}+1}{2})<\pi \end{matrix}\right,$$

y uchun yechish

$$\left\{\begin{matrix}y=\frac{\sqrt{4x+1}+1}{2}\text{, }&x\geq -\frac{1}{4}\\y=\frac{-\sqrt{4x+1}+1}{2}\text{, }&x\geq -\frac{1}{4}\text{ and }x\leq 0\end{matrix}\right,$$