r=sqrt{(r-2)^2+A^2} eching

$$r = \sqrt { ( r - 2 ) ^ { 2 } + A ^ { 2 } }$$

r uchun yechish

$$r=\frac{A^{2}+4}{4}$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$r^{2}=\left(\sqrt{\left(r-2\right)^{2}+A^{2}}\right)^{2}$$

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

$$r^{2}=\left(\sqrt{r^{2}-4r+4+A^{2}}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{r^{2}-4r+4+A^{2}} ga hisoblang va r^{2}-4r+4+A^{2} ni qiymatni oling.

$$r^{2}=r^{2}-4r+4+A^{2}$$

Ikkala tarafdan r^{2} ni ayirish.

$$r^{2}-r^{2}=-4r+4+A^{2}$$

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

$$0=-4r+4+A^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-4r+4+A^{2}=0$$

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-4r+A^{2}=-4$$

Ikkala tarafdan A^{2} ni ayirish.

$$-4r=-4-A^{2}$$

Tenglama standart shaklda.

$$-4r=-A^{2}-4$$

Ikki tarafini -4 ga bo‘ling.

$$\frac{-4r}{-4}=\frac{-A^{2}-4}{-4}$$

-4 ga bo'lish -4 ga ko'paytirishni bekor qiladi.

$$r=\frac{-A^{2}-4}{-4}$$

-4-A^{2} ni -4 ga bo'lish.

$$r=\frac{A^{2}}{4}+1$$

r=\sqrt{\left(r-2\right)^{2}+A^{2}} tenglamasida r uchun \frac{A^{2}}{4}+1 ni almashtiring.

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

Qisqartirish. r=\frac{A^{2}}{4}+1 tenglamani qoniqtiradi.

$$\frac{1}{4}A^{2}+1=\frac{1}{4}\left(16+8A^{2}+A^{4}\right)^{\frac{1}{2}}$$

r=\sqrt{\left(r-2\right)^{2}+A^{2}} tenglamasi noyob yechimga ega.

$$r=\frac{A^{2}}{4}+1$$

A uchun yechish

$$A=2\sqrt{r-1}$$

$$A=-2\sqrt{r-1}\text{, }r\geq 1$$