y=sqrt{2x+2}+1 eching

$$y = \sqrt { 2 x + 2 } + 1$$

x uchun yechish

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

$$y-1\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{2x+2}+1=y$$

Ikkala tarafdan 1 ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

\left(y-1\right)^{2}-2 ni 2 ga bo'lish.

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

x uchun yechish (complex solution)

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

$$y=1\text{ or }arg(1-y)\geq \pi$$

y uchun yechish (complex solution)

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

y uchun yechish

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

$$x\geq -1$$