Question

Решите пожалуйста В1 с обеих вариантов

Answer 1

В1 вариант 1.

$(\frac{1}{(b-y)(y-5)}-\frac{1}{(b-y)(b-5)}-\frac{1}{(b-5)(y-5)})*\frac{b^2-9y^2}{b^4+y^4}=0\\\\1)\frac{1}{(b-y)(y-5)}-\frac{1}{(b-y)(b-5)}-\frac{1}{(b-5)(y-5)}=\\\\=\frac{(b-5)-(y-5)-(b-y)}{(b-y)(y-5)(b-5)}=\frac{0}{(b-y)(y-5)(b-5)}=0\\\\2) 0 * \frac{b^2-9y^2}{b^4+y^4}=0$

Ответ: 0.

В1 вариант 2.

$(\frac{1}{(a-x)(x-1)}-\frac{1}{(a-x)(a-1)}-\frac{1}{(a-1)(x-1)})*\frac{a^3-8x^3}{a^4+b^4}=0\\\\1)\frac{1}{(a-x)(x-1)}-\frac{1}{(a-x)(a-1)}-\frac{1}{(a-1)(x-1)}=\\\\=\frac{(a-1)-(x-1)-(a-x)}{(a-x)(a-1)(x-1)}=\frac{0}{(a-x)(a-1)(x-1)}=0\\\\2)0*\frac{a^3-8x^3}{a^4+b^4}=0$

Ответ: 0.