b^2-(5-x)^2=5^2-x^2 eching

$$b ^ { 2 } - ( 5 - x ) ^ { 2 } = 5 ^ { 2 } - x ^ { 2 }$$

x uchun yechish

$$x=-\frac{b^{2}}{10}+5$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5-x\right)^{2} kengaytirilishi uchun ishlating.

$$b^{2}-\left(25-10x+x^{2}\right)=5^{2}-x^{2}$$

25-10x+x^{2} teskarisini topish uchun har birining teskarisini toping.

$$b^{2}-25+10x-x^{2}=5^{2}-x^{2}$$

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$b^{2}-25+10x-x^{2}=25-x^{2}$$

x^{2} ni ikki tarafga qo’shing.

$$b^{2}-25+10x-x^{2}+x^{2}=25$$

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

$$b^{2}-25+10x=25$$

Ikkala tarafdan b^{2} ni ayirish.

$$-25+10x=25-b^{2}$$

25 ni ikki tarafga qo’shing.

$$10x=25-b^{2}+25$$

50 olish uchun 25 va 25'ni qo'shing.

$$10x=50-b^{2}$$

Ikki tarafini 10 ga bo‘ling.

$$\frac{10x}{10}=\frac{50-b^{2}}{10}$$

10 ga bo'lish 10 ga ko'paytirishni bekor qiladi.

$$x=\frac{50-b^{2}}{10}$$

50-b^{2} ni 10 ga bo'lish.

$$x=-\frac{b^{2}}{10}+5$$

b uchun yechish (complex solution)

$$b=-\sqrt{50-10x}$$

$$b=\sqrt{50-10x}$$

b uchun yechish

$$b=\sqrt{50-10x}$$

$$b=-\sqrt{50-10x}\text{, }x\leq 5$$