2(2-z)=xsqrt{left(2-zright)^2+4} eching

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

x uchun yechish

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

2 ga 2-z ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

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

$$4-2z=x\sqrt{4-4z+z^{2}+4}$$

8 olish uchun 4 va 4'ni qo'shing.

$$4-2z=x\sqrt{8-4z+z^{2}}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x\sqrt{8-4z+z^{2}}=4-2z$$

Tenglama standart shaklda.

$$\sqrt{z^{2}-4z+8}x=4-2z$$

Ikki tarafini \sqrt{8-4z+z^{2}} ga bo‘ling.

$$\frac{\sqrt{z^{2}-4z+8}x}{\sqrt{z^{2}-4z+8}}=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

\sqrt{8-4z+z^{2}} ga bo'lish \sqrt{8-4z+z^{2}} ga ko'paytirishni bekor qiladi.

$$x=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

4-2z ni \sqrt{8-4z+z^{2}} ga bo'lish.

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

z uchun yechish

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

$$x>-2\text{ and }x<2$$