(5x+b)(cx+d)=10x^2-c eching

$$( 5 x + b ) ( c x + d ) = 10 x ^ { 2 } - c$$

b uchun yechish

$$\left\{\begin{matrix}b=-\frac{5cx^{2}-10x^{2}+5dx+c}{cx+d}\text{, }&d\neq -cx\\b\in \mathrm{R}\text{, }&c=\sqrt[3]{10}\times 10^{\frac{2}{3}}x^{2}\text{ and }d=-10x^{3}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

5x+b ga cx+d ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5cx^{2}+5xd+bcx+bd=10x^{2}-c$$

Ikkala tarafdan 5cx^{2} ni ayirish.

$$5xd+bcx+bd=10x^{2}-c-5cx^{2}$$

Ikkala tarafdan 5xd ni ayirish.

$$bcx+bd=10x^{2}-c-5cx^{2}-5xd$$

b'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(cx+d\right)b=10x^{2}-c-5cx^{2}-5xd$$

Tenglama standart shaklda.

$$\left(cx+d\right)b=-5cx^{2}+10x^{2}-5dx-c$$

Ikki tarafini cx+d ga bo‘ling.

$$\frac{\left(cx+d\right)b}{cx+d}=\frac{-5cx^{2}+10x^{2}-5dx-c}{cx+d}$$

cx+d ga bo'lish cx+d ga ko'paytirishni bekor qiladi.

$$b=\frac{-5cx^{2}+10x^{2}-5dx-c}{cx+d}$$