Question

СРОЧНО ДАЮ 100 БАЛІВ АЛГЕБРА 8 КЛАС

Answer 1

Объяснение:

$\displaystyle\\1.\\\\x+\frac{8}{x^2} =\frac{x*x^2+8}{x^2} =\frac{x^3+8}{x^2}.\\\\2.\\\\\frac{2}{8+x^3}-\frac{1}{4-x^2}=\frac{2}{x^3+8}+\frac{1}{x^2-4}= \frac{2*(x^2-4)+1*(x^3+8)}{(x^3+8)*(x^2-4)}=\\\\\\ =\frac{2x^2-8+x^3+8}{(x^3+8)*(x^2-4)}=\frac{x^3+2x^2}{(x^3+8)*(x^2-2^2)}=\frac{x^2*(x+2)}{(x^3+8)*(x+2)*(x-2)}=\\\\\\ =\frac{x^2}{(x^3+8)*(x-2)}.$

$\displaystyle\\ 3.\\\\\frac{x^3+8}{x^2}*\frac{x^2}{(x^3+8)*(x-2)}=\frac{1}{x-2} .\\\\ 4.\\\\\frac{1}{x-2} -\frac{2+x-x^2}{x^2-4x+4} =\frac{1}{x-2} +\frac{x^2-x-2}{(x-2)^2}=\frac{x-2+x^2-x-2}{(x-2)^2}=\\\\\\=\frac{x^2-4}{(x-2)^2}=\frac{(x-2)*(x+2)}{(x-2)^2} =\frac{x+2}{x-2}.$

Answer 2

Ответ:

Доказать тождество . Применяем формулы сокращённого умножения .

$\bf \displaystyle \Big(x+\frac{8}{x^2}\Big)\cdot \Big(\frac{2}{8+x^3}-\frac{1}{4-x^2}\Big)-\frac{2+x-x^2}{x^2-4x+4}=\frac{x+2}{x-2}\\\\\\{}\ \ \ \displaystyle \Big(x+\frac{8}{x^2}\Big)\cdot \Big(\frac{2}{8+x^3}-\frac{1}{4-x^2}\Big)-\frac{2+x-x^2}{x^2-4x+4}=\\\\\\=\frac{x^3+8}{x^2}\cdot \frac{8-2x^2-8-x^3}{(8+x^3)(4-x^2)}-\frac{2+x-x^2}{(x-2)^2}=\\\\\\=\frac{-2x^2-x^3}{x^2\, (2-x)(2+x)}-\frac{2+x-x^2}{(x-2)^2}=\frac{-x^2\, (2+x)}{x^2\, (2-x)(2+x)}-\frac{2+x-x^2}{(x-2)^2}=$    

 $\bf \displaystyle =\frac{-1}{2-x}-\frac{2+x-x^2}{(x-2)^2}=\frac{1}{x-2}-\frac{2+x-x^2}{(x-2)^2}=\frac{x-2-2-x+x^2}{(x-2)^2}=\\\\\\=\frac{x^2-4}{(x-2)^2}=\frac{(x-2)(x+2)}{(x-2)^2}=\frac{x+2}{x-2}$  

$\bf \dfrac{x+2}{x-2}=\dfrac{x+2}{x-2}$              

Тождество доказано .