Question

* РЕШИТЕ! 50 БАЛЛОВ !*

Answer 1

Ответ:

$4)\ \ \dfrac{\dfrac{3x-y}{y}+1}{\dfrac{3x+y}{y}-1}\ +\ \dfrac{3-\dfrac{y}{x}}{\dfrac{3x}{y}-1}\ =\ \dfrac{\dfrac{3x-y+y}{y}}{\dfrac{3x+y-y}{y}}\ +\ \dfrac{\dfrac{3x-y}{y}}{\dfrac{3x-y}{y}}=\\\\\\=\dfrac{3x}{3x}+\dfrac{3x-y}{3x-y}=1+1=2$

$5)\ \ \Big(\dfrac{1}{(a-x)(x-1)}-\dfrac{1}{(a-x)(a-1)}-\dfrac{1}{(a-1)(x-1)}\Big)\cdot \dfrac{a^3-8x^3}{a^4+b^4}=\\\\\\=\dfrac{(a-1)-(x-1)-(a-x)}{(a-x)(x-1)(a-1)}\cdot \dfrac{(a-2x)(a^2+2ax+4x^2)}{a^4+b^4}=\\\\\\=\dfrac{a-1-x+1-a+x}{(a-x)(x-1)(a-1)}\cdot \dfrac{(a-2x)(a^2+2ax+4x^2)}{a^4+b^4}=\\\\\\=\dfrac{0}{(a-x)(x-1)(a-1)}\cdot \dfrac{(a-2x)(a^2+2ax+4x^2)}{a^4+b^4}=\\\\\\=0\cdot \dfrac{(a-2x)(a^2+2ax+4x^2)}{a^4+b^4}\, =\, 0$