sqrt{x+3}+2=sqrt{x+11} eching

$$\sqrt { x + 3 } + 2 = \sqrt { x + 11 }$$

x uchun yechish

$$x=-2$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{x+3}+2\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

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

$$\left(\sqrt{x+3}\right)^{2}+4\sqrt{x+3}+4=\left(\sqrt{x+11}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+3} ga hisoblang va x+3 ni qiymatni oling.

$$x+3+4\sqrt{x+3}+4=\left(\sqrt{x+11}\right)^{2}$$

7 olish uchun 3 va 4'ni qo'shing.

$$x+7+4\sqrt{x+3}=\left(\sqrt{x+11}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+11} ga hisoblang va x+11 ni qiymatni oling.

$$x+7+4\sqrt{x+3}=x+11$$

Ikkala tarafdan x ni ayirish.

$$x+7+4\sqrt{x+3}-x=11$$

0 ni olish uchun x va -x ni birlashtirish.

$$7+4\sqrt{x+3}=11$$

Ikkala tarafdan 7 ni ayirish.

$$4\sqrt{x+3}=11-7$$

4 olish uchun 11 dan 7 ni ayirish.

$$4\sqrt{x+3}=4$$

Ikki tarafini 4 ga bo‘ling.

$$\sqrt{x+3}=\frac{4}{4}$$

1 ni olish uchun 4 ni 4 ga bo‘ling.

$$\sqrt{x+3}=1$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$x+3=1$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x+3-3=1-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$x=1-3$$

1 dan 3 ni ayirish.

$$x=-2$$

\sqrt{x+3}+2=\sqrt{x+11} tenglamasida x uchun -2 ni almashtiring.

$$\sqrt{-2+3}+2=\sqrt{-2+11}$$

Qisqartirish. x=-2 tenglamani qoniqtiradi.

$$3=3$$

\sqrt{x+3}+2=\sqrt{x+11} tenglamasi noyob yechimga ega.

$$x=-2$$