y=1/2sqrt{x+1}-5 eching

$$y = \frac { 1 } { 2 } \sqrt { x + 1 } - 5$$

x uchun yechish

$$x=\left(2y+9\right)\left(2y+11\right)$$

$$2y+10\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{1}{2}\sqrt{x+1}-5=y$$

5 ni ikki tarafga qo’shing.

$$\frac{1}{2}\sqrt{x+1}=y+5$$

Ikkala tarafini 2 ga ko‘paytiring.

$$\frac{\frac{1}{2}\sqrt{x+1}}{\frac{1}{2}}=\frac{y+5}{\frac{1}{2}}$$

\frac{1}{2} ga bo'lish \frac{1}{2} ga ko'paytirishni bekor qiladi.

$$\sqrt{x+1}=\frac{y+5}{\frac{1}{2}}$$

y+5 ni \frac{1}{2} ga bo'lish y+5 ga k'paytirish \frac{1}{2} ga qaytarish.

$$\sqrt{x+1}=2y+10$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$x+1=4\left(y+5\right)^{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x+1-1=4\left(y+5\right)^{2}-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$x=4\left(y+5\right)^{2}-1$$

4\left(5+y\right)^{2} dan 1 ni ayirish.

$$x=4y^{2}+40y+99$$

x uchun yechish (complex solution)

$$x=\left(2y+9\right)\left(2y+11\right)$$

$$y=-5\text{ or }arg(2y+10)<\pi$$

y uchun yechish (complex solution)

$$y=\frac{\sqrt{x+1}-10}{2}$$

y uchun yechish

$$y=\frac{\sqrt{x+1}-10}{2}$$

$$x\geq -1$$