x+2/(x+5)(x+3)=A/(x+5)+B/(x+3) eching

$$\frac { x + 2 } { ( x + 5 ) ( x + 3 ) } = \frac { A } { ( x + 5 ) } + \frac { B } { ( x + 3 ) }$$

A uchun yechish

$$A=-\frac{Bx-x+5B-2}{x+3}$$

$$x\neq -5\text{ and }x\neq -3$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \left(x+3\right)\left(x+5\right) ga, \left(x+5\right)\left(x+3\right),x+5,x+3 ning eng kichik karralisiga ko‘paytiring.

$$x+2=\left(x+3\right)A+\left(x+5\right)B$$

x+3 ga A ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+2=xA+3A+\left(x+5\right)B$$

x+5 ga B ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+2=xA+3A+xB+5B$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$xA+3A+xB+5B=x+2$$

Ikkala tarafdan xB ni ayirish.

$$xA+3A+5B=x+2-xB$$

Ikkala tarafdan 5B ni ayirish.

$$xA+3A=x+2-xB-5B$$

A'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x+3\right)A=x+2-xB-5B$$

Tenglama standart shaklda.

$$\left(x+3\right)A=2-5B+x-Bx$$

Ikki tarafini x+3 ga bo‘ling.

$$\frac{\left(x+3\right)A}{x+3}=\frac{2-5B+x-Bx}{x+3}$$

x+3 ga bo'lish x+3 ga ko'paytirishni bekor qiladi.

$$A=\frac{2-5B+x-Bx}{x+3}$$

B uchun yechish

$$B=-\frac{Ax-x+3A-2}{x+5}$$

$$x\neq -5\text{ and }x\neq -3$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \left(x+3\right)\left(x+5\right) ga, \left(x+5\right)\left(x+3\right),x+5,x+3 ning eng kichik karralisiga ko‘paytiring.

$$x+2=\left(x+3\right)A+\left(x+5\right)B$$

x+3 ga A ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+2=xA+3A+\left(x+5\right)B$$

x+5 ga B ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+2=xA+3A+xB+5B$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$xA+3A+xB+5B=x+2$$

Ikkala tarafdan xA ni ayirish.

$$3A+xB+5B=x+2-xA$$

Ikkala tarafdan 3A ni ayirish.

$$xB+5B=x+2-xA-3A$$

B'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x+5\right)B=x+2-xA-3A$$

Tenglama standart shaklda.

$$\left(x+5\right)B=2-3A+x-Ax$$

Ikki tarafini x+5 ga bo‘ling.

$$\frac{\left(x+5\right)B}{x+5}=\frac{2-3A+x-Ax}{x+5}$$

x+5 ga bo'lish x+5 ga ko'paytirishni bekor qiladi.

$$B=\frac{2-3A+x-Ax}{x+5}$$