Question

СРОЧНО НУЖНА ПОМОЩЬ!!!!! ЗАРАНЕЕ СПАСИБО


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Answer 1

Объясн$\frac{x^2+5}{(x-5)^2}:(\frac{5}{x+5}+\frac{x^2+25}{x^2-25}-\frac{5}{5-x})=\frac{x(x+5)}{(x-5)^2}:(\frac{5}{x+5}+\frac{x^2+25}{(x-5)(x+5)}-\frac{5}{-(x-5)}=\frac{x(x+5)}{(x-5)^2}:(\frac{5}{x+5}+\frac{x^2+25}{(x-5)(x+5)}+\frac{5}{x-5})=\frac{x(x+5)}{(x-5)^2}:\frac{5(x-5)+x^2+25+5(x+5)}{(x-5)(x+5)}=\frac{x(x+5)}{(x-5)^2}:\frac{5x-25+x^2+25+5x+25}{(x-5)(x+5)}=\frac{x(x+5)}{(x-5)^2}:\frac{10+x^2+25}{(x-5)(x+5)}=\frac{x(x+5)}{(x-5)^2}:\frac{x^2+10x+25}{(x-5)(x+5)}=\frac{x(x+5)}{(x-5)^2}:$$\frac{(x+5)^2}{(x-5)(x+5)}=\frac{x(x+5)}{(x-5)^2}:\frac{x+5}{x-5}=\frac{x(x+5)}{(x-5)^2}*\frac{x-5}{x+5}=\frac{x}{x-5}$

$\frac{a^3+27}{a-1}*(\frac{a-3}{a^2-3a+9}+\frac{a+9}{a^3+27}):\frac{a^2+a}{a^2-1}=\frac{(a+3)(a^2-3a+9)}{a-1}*(\frac{a-3}{a^2-3a+9}+\frac{a+9}{(a+3)(a^2-3a+9)})*\frac{a^2-1}{a^2+a}=\frac{(a+3)(a^2-3a+9)}{a-1}*\frac{(a+3)(a-3)+a+9}{(a+3)(a^2-3a+9)}*\frac{(a-1)(a+1)}{a^2+a}=\frac{1}{a-1}*(a^2-9+a+9)*\frac{(a-1)(a+1)}{a^2+a}=\frac{1}{a-1}*(a^2+a)*\frac{(a-1)(a+1)}{a^2+a}=a+1$