y=sqrt{3x-12} eching

$$y = \sqrt { 3 x - 12 }$$

x uchun yechish

$$x=\frac{y^{2}+12}{3}$$

$$y\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{3x-12}=y$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$3x-12=y^{2}$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$3x-12-\left(-12\right)=y^{2}-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$3x=y^{2}-\left(-12\right)$$

y^{2} dan -12 ni ayirish.

$$3x=y^{2}+12$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x}{3}=\frac{y^{2}+12}{3}$$

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

$$x=\frac{y^{2}+12}{3}$$

y^{2}+12 ni 3 ga bo'lish.

$$x=\frac{y^{2}}{3}+4$$

x uchun yechish (complex solution)

$$x=\frac{y^{2}+12}{3}$$

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

y uchun yechish (complex solution)

$$y=\sqrt{3\left(x-4\right)}$$

y uchun yechish

$$y=\sqrt{3\left(x-4\right)}$$

$$x\geq 4$$