sqrt{x^2-1}=x+2 eching

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

x uchun yechish

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1,25$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{x^{2}-1} ga hisoblang va x^{2}-1 ni qiymatni oling.

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

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

$$x^{2}-1=x^{2}+4x+4$$

Ikkala tarafdan x^{2} ni ayirish.

$$x^{2}-1-x^{2}=4x+4$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$-1=4x+4$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$4x+4=-1$$

Ikkala tarafdan 4 ni ayirish.

$$4x=-1-4$$

-5 olish uchun -1 dan 4 ni ayirish.

$$4x=-5$$

Ikki tarafini 4 ga bo‘ling.

$$x=\frac{-5}{4}$$

\frac{-5}{4} kasri manfiy belgini olib tashlash bilan -\frac{5}{4} sifatida qayta yozilishi mumkin.

$$x=-\frac{5}{4}$$

\sqrt{x^{2}-1}=x+2 tenglamasida x uchun -\frac{5}{4} ni almashtiring.

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

Qisqartirish. x=-\frac{5}{4} tenglamani qoniqtiradi.

$$\frac{3}{4}=\frac{3}{4}$$

\sqrt{x^{2}-1}=x+2 tenglamasi noyob yechimga ega.

$$x=-\frac{5}{4}$$