sqrt{a^2-b^2}=a-b eching

$$\sqrt { a ^ { 2 } - b ^ { 2 } } = a - b$$

a uchun yechish (complex solution)

$$\left\{\begin{matrix}\\a=b\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&b=0\text{ and }arg(a)<\pi \end{matrix}\right,$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{a^{2}-b^{2}}\right)^{2}=\left(a-b\right)^{2}$$

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

$$a^{2}-b^{2}=\left(a-b\right)^{2}$$

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

$$a^{2}-b^{2}=a^{2}-2ab+b^{2}$$

Ikkala tarafdan a^{2} ni ayirish.

$$a^{2}-b^{2}-a^{2}=-2ab+b^{2}$$

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

$$-b^{2}=-2ab+b^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-2ab+b^{2}=-b^{2}$$

Ikkala tarafdan b^{2} ni ayirish.

$$-2ab=-b^{2}-b^{2}$$

-2b^{2} ni olish uchun -b^{2} va -b^{2} ni birlashtirish.

$$-2ab=-2b^{2}$$

-2ni ikki tarafidan bekor qilish.

$$ab=b^{2}$$

Tenglama standart shaklda.

$$ba=b^{2}$$

Ikki tarafini b ga bo‘ling.

$$\frac{ba}{b}=\frac{b^{2}}{b}$$

b ga bo'lish b ga ko'paytirishni bekor qiladi.

$$a=\frac{b^{2}}{b}$$

b^{2} ni b ga bo'lish.

$$a=b$$

\sqrt{a^{2}-b^{2}}=a-b tenglamasida a uchun b ni almashtiring.

$$\sqrt{b^{2}-b^{2}}=b-b$$

Qisqartirish. a=b tenglamani qoniqtiradi.

$$0=0$$

\sqrt{a^{2}-b^{2}}=a-b tenglamasi noyob yechimga ega.

$$a=b$$

a uchun yechish

$$\left\{\begin{matrix}a=b\text{, }&b\neq 0\\a\geq 0\text{, }&b=0\end{matrix}\right,$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{a^{2}-b^{2}}\right)^{2}=\left(a-b\right)^{2}$$

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

$$a^{2}-b^{2}=\left(a-b\right)^{2}$$

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

$$a^{2}-b^{2}=a^{2}-2ab+b^{2}$$

Ikkala tarafdan a^{2} ni ayirish.

$$a^{2}-b^{2}-a^{2}=-2ab+b^{2}$$

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

$$-b^{2}=-2ab+b^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-2ab+b^{2}=-b^{2}$$

Ikkala tarafdan b^{2} ni ayirish.

$$-2ab=-b^{2}-b^{2}$$

-2b^{2} ni olish uchun -b^{2} va -b^{2} ni birlashtirish.

$$-2ab=-2b^{2}$$

-2ni ikki tarafidan bekor qilish.

$$ab=b^{2}$$

Tenglama standart shaklda.

$$ba=b^{2}$$

Ikki tarafini b ga bo‘ling.

$$\frac{ba}{b}=\frac{b^{2}}{b}$$

b ga bo'lish b ga ko'paytirishni bekor qiladi.

$$a=\frac{b^{2}}{b}$$

b^{2} ni b ga bo'lish.

$$a=b$$

\sqrt{a^{2}-b^{2}}=a-b tenglamasida a uchun b ni almashtiring.

$$\sqrt{b^{2}-b^{2}}=b-b$$

Qisqartirish. a=b tenglamani qoniqtiradi.

$$0=0$$

\sqrt{a^{2}-b^{2}}=a-b tenglamasi noyob yechimga ega.

$$a=b$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}\\b=a\text{, }&\text{unconditionally}\\b=0\text{, }&arg(a)<\pi \end{matrix}\right,$$

b uchun yechish

$$\left\{\begin{matrix}\\b=a\text{, }&\text{unconditionally}\\b=0\text{, }&a\geq 0\end{matrix}\right,$$