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<span>2log5(x-1)=log 5(12x+1)
</span><span>log5(x-1)^2=log 5(12x+1)
Т.к. основания равны,то:
(x-1)^2=12x+1
x^2-2x+1-12x-1=0
x^2-14x=0
x(x-7)=0
x=0 или x-7=0
х=7
ОДЗ:x-1>0
x>1
__________
12x+1>0
12x>-1
x>-1/12
Ответ:x=7
(x=0 не подходит по одз)</span>
1)25y^2 - a^2 = (5y - a)(5y + a)
2)c^2 + 4bc + 4b^2 = ( c+2b)(c+2b)