sqrt{y}+sqrt{y+2}=3 eching

$$\sqrt{ y } + \sqrt{ y+2 } =3$$

y uchun yechish

$$y = \frac{49}{36} = 1\frac{13}{36} \approx 1,361111111$$

Yechim qadamlari

Tenglamaning ikkala tarafidan \sqrt{y+2} ni ayirish.

$$\sqrt{y}=3-\sqrt{y+2}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{y}\right)^{2}=\left(3-\sqrt{y+2}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{y} ga hisoblang va y ni qiymatni oling.

$$y=\left(3-\sqrt{y+2}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3-\sqrt{y+2}\right)^{2} kengaytirilishi uchun ishlating.

$$y=9-6\sqrt{y+2}+\left(\sqrt{y+2}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{y+2} ga hisoblang va y+2 ni qiymatni oling.

$$y=9-6\sqrt{y+2}+y+2$$

11 olish uchun 9 va 2'ni qo'shing.

$$y=11-6\sqrt{y+2}+y$$

6\sqrt{y+2} ni ikki tarafga qo’shing.

$$y+6\sqrt{y+2}=11+y$$

Ikkala tarafdan y ni ayirish.

$$y+6\sqrt{y+2}-y=11$$

0 ni olish uchun y va -y ni birlashtirish.

$$6\sqrt{y+2}=11$$

Ikki tarafini 6 ga bo‘ling.

$$\sqrt{y+2}=\frac{11}{6}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$y+2=\frac{121}{36}$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$y+2-2=\frac{121}{36}-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$y=\frac{121}{36}-2$$

\frac{121}{36} dan 2 ni ayirish.

$$y=\frac{49}{36}$$

\sqrt{y}+\sqrt{y+2}=3 tenglamasida y uchun \frac{49}{36} ni almashtiring.

$$\sqrt{\frac{49}{36}}+\sqrt{\frac{49}{36}+2}=3$$

Qisqartirish. y=\frac{49}{36} tenglamani qoniqtiradi.

$$3=3$$

\sqrt{y}=-\sqrt{y+2}+3 tenglamasi noyob yechimga ega.

$$y=\frac{49}{36}$$