sqrt{2-x}=x-2/2 eching

$$\sqrt{ 2-x } = \frac{ x-2 }{ 2 }$$

x uchun yechish

$$x=2$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

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

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

\frac{x-2}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

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

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

$$2-x=\frac{x^{2}-4x+4}{2^{2}}$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$2-x=\frac{x^{2}-4x+4}{4}$$

\frac{1}{4}x^{2}-x+1 natijani olish uchun x^{2}-4x+4 ning har bir ifodasini 4 ga bo‘ling.

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

Ikkala tarafdan \frac{1}{4}x^{2} ni ayirish.

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

x ni ikki tarafga qo’shing.

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

0 ni olish uchun -x va x ni birlashtirish.

$$2-\frac{1}{4}x^{2}=1$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{4}x^{2}=1-2$$

-1 olish uchun 1 dan 2 ni ayirish.

$$-\frac{1}{4}x^{2}=-1$$

Ikki tarafini -4 va teskari kasri -\frac{1}{4} ga ko‘paytiring.

$$x^{2}=-\left(-4\right)$$

4 hosil qilish uchun -1 va -4 ni ko'paytirish.

$$x^{2}=4$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$x=2 x=-2$$

\sqrt{2-x}=\frac{x-2}{2} tenglamasida x uchun 2 ni almashtiring.

$$\sqrt{2-2}=\frac{2-2}{2}$$

Qisqartirish. x=2 tenglamani qoniqtiradi.

$$0=0$$

\sqrt{2-x}=\frac{x-2}{2} tenglamasida x uchun -2 ni almashtiring.

$$\sqrt{2-\left(-2\right)}=\frac{-2-2}{2}$$

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

$$2=-2$$

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

$$x=2$$