решите, пожалуйста:а²-4/3а+1 × (3а-1)(3а+1)/(3а²+5а-2). Created by liiiluuuli. algebra-ru

(3а²+5а-2)...

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решите, пожалуйста:
а²-4/3а+1 × (3а-1)(3а+1)/(3а²+5а-2)

Answers & Comments

армения20171 (à²-4)/(3a+1)*(3a-1)(3a+1)/(3a²+5a-2)
=(a²-4)(3a-1)/(3a²+5a-2)
3a²+5a-2=0
D=25+24=49=7²
a=(-5±7)/6
a1=-2;a2=1/3
3a²+5a-2=3(a-1/3)(a+2)
(a-2)(a+2)•1/3(a-1/3)/3(a+2)(a-1/3)=
1/3(a-2):3==1/6(a-2)=(a-2)/6

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