x=sqrt{3y-1/2} eching

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

y uchun yechish

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

$$x\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$3y-\frac{1}{2}-\left(-\frac{1}{2}\right)=x^{2}-\left(-\frac{1}{2}\right)$$

O‘zidan -\frac{1}{2} ayirilsa 0 qoladi.

$$3y=x^{2}-\left(-\frac{1}{2}\right)$$

x^{2} dan -\frac{1}{2} ni ayirish.

$$3y=x^{2}+\frac{1}{2}$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3y}{3}=\frac{x^{2}+\frac{1}{2}}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$y=\frac{x^{2}+\frac{1}{2}}{3}$$

x^{2}+\frac{1}{2} ni 3 ga bo'lish.

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{12y-2}}{2}$$

y uchun yechish (complex solution)

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

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

x uchun yechish

$$x=\frac{\sqrt{12y-2}}{2}$$

$$y\geq \frac{1}{6}$$