x+1=sqrt{2x+5} eching

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

x uchun yechish

$$x=2$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

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

2 daraja ko‘rsatkichini \sqrt{2x+5} ga hisoblang va 2x+5 ni qiymatni oling.

$$x^{2}+2x+1=2x+5$$

Ikkala tarafdan 2x ni ayirish.

$$x^{2}+2x+1-2x=5$$

0 ni olish uchun 2x va -2x ni birlashtirish.

$$x^{2}+1=5$$

Ikkala tarafdan 5 ni ayirish.

$$x^{2}+1-5=0$$

-4 olish uchun 1 dan 5 ni ayirish.

$$x^{2}-4=0$$

Hisoblang: x^{2}-4. x^{2}-4 ni x^{2}-2^{2} sifatida qaytadan yozish. Kvadratlarning farqini ushbu formula bilan hisoblash mumkin: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

Tenglamani yechish uchun x-2=0 va x+2=0 ni yeching.

$$x=2 x=-2$$

x+1=\sqrt{2x+5} tenglamasida x uchun 2 ni almashtiring.

$$2+1=\sqrt{2\times 2+5}$$

Qisqartirish. x=2 tenglamani qoniqtiradi.

$$3=3$$

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

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

Qisqartirish. x=-2 qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$-1=1$$

x+1=\sqrt{2x+5} tenglamasi noyob yechimga ega.

$$x=2$$