1)x²+10x+25=(x+5)², √(x+5)²=/x+5/, -5∠x ⇒/x+5/=x+5
35(x+5)/(x+5)=35
Otvet:35
2)x∠0, /x-1/=-(x-1)
-(x-1)/309x(x-1)=-1/309x , x≠0
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Answers & Comments
1)x²+10x+25=(x+5)², √(x+5)²=/x+5/, -5∠x ⇒/x+5/=x+5
35(x+5)/(x+5)=35
Otvet:35
2)x∠0, /x-1/=-(x-1)
-(x-1)/309x(x-1)=-1/309x , x≠0