a+1/b=a-1/b+b+1/a eching

$$\frac { a + 1 } { b } = \frac { a - 1 } { b } + \frac { b + 1 } { a }$$

a uchun yechish

$$a=\frac{b\left(b+1\right)}{2}$$

$$b\neq -1\text{ and }b\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

a qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini ab ga, b,a ning eng kichik karralisiga ko‘paytiring.

$$a\left(a+1\right)=a\left(a-1\right)+b\left(b+1\right)$$

a ga a+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$a^{2}+a=a\left(a-1\right)+b\left(b+1\right)$$

a ga a-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$a^{2}+a=a^{2}-a+b\left(b+1\right)$$

b ga b+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Ikkala tarafdan a^{2} ni ayirish.

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

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

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

a ni ikki tarafga qo’shing.

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

2a ni olish uchun a va a ni birlashtirish.

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

Ikki tarafini 2 ga bo‘ling.

$$\frac{2a}{2}=\frac{b\left(b+1\right)}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$a=\frac{b\left(b+1\right)}{2}$$

a qiymati 0 teng bo‘lmaydi.

$$a=\frac{b\left(b+1\right)}{2}\text{, }a\neq 0$$

b uchun yechish

$$b=\frac{-\sqrt{8a+1}-1}{2}$$

$$b=\frac{\sqrt{8a+1}-1}{2}\text{, }a\neq 0\text{ and }a\geq -\frac{1}{8}$$