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

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

x uchun yechish

$$x=4$$

Yechim qadamlari

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

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

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

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

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

Ikkala tarafdan x^{2} ni ayirish.

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

2x ni ikki tarafga qo’shing.

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

4x ni olish uchun 2x va 2x ni birlashtirish.

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

Ikkala tarafdan 1 ni ayirish.

$$4x+1-x^{2}-1=0$$

0 olish uchun 1 dan 1 ni ayirish.

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

x omili.

$$x\left(4-x\right)=0$$

Tenglamani yechish uchun x=0 va 4-x=0 ni yeching.

$$x=0 x=4$$

\sqrt{2x+1}+1=x tenglamasida x uchun 0 ni almashtiring.

$$\sqrt{2\times 0+1}+1=0$$

Qisqartirish. x=0 qiymati bu tenglamani qoniqtirmaydi.

$$2=0$$

\sqrt{2x+1}+1=x tenglamasida x uchun 4 ni almashtiring.

$$\sqrt{2\times 4+1}+1=4$$

Qisqartirish. x=4 tenglamani qoniqtiradi.

$$4=4$$

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

$$x=4$$