a^2+b^2/ab=a+c/b eching

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

a uchun yechish

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

$$b\neq 0\text{ and }c\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

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

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

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

Ikkala tarafdan a^{2} ni ayirish.

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

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

$$b^{2}=ac$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ac=b^{2}$$

Tenglama standart shaklda.

$$ca=b^{2}$$

Ikki tarafini c ga bo‘ling.

$$\frac{ca}{c}=\frac{b^{2}}{c}$$

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

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

a qiymati 0 teng bo‘lmaydi.

$$a=\frac{b^{2}}{c}\text{, }a\neq 0$$

b uchun yechish (complex solution)

$$b=-\sqrt{a}\sqrt{c}$$

$$b=\sqrt{a}\sqrt{c}\text{, }a\neq 0\text{ and }c\neq 0$$

b uchun yechish

$$b=\sqrt{ac}$$

$$b=-\sqrt{ac}\text{, }\left(c<0\text{ and }a<0\right)\text{ or }\left(a>0\text{ and }c>0\right)$$