log4 (x+3)+log(1/4) (x-3)=log(1/4) (x-1)+log4 (2x+1)的解集,过程?
2个回答
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log4(x+3)+log(1/4)(x-3)=log(1/4)(x-1)+log4(2x+1)
log4(x+3)-log4(2x+1)=log(1/4)(x-1)-log(1/4)(x-3)
log4(x+3)/(2x+1)=log(1/4)(x-1)/(x-3)
log4(x+3)/(2x+1)=log4(x-3)/(x-1)
(x+3)/(2x+1)=(x-3)/(x-1)
(2x+2)(x-3)=(x+3)(x-1)
2x2-6x+2x-6=x2+2x-3
x2-6x-3=0
x=3±2√3
log4(x+3)-log4(2x+1)=log(1/4)(x-1)-log(1/4)(x-3)
log4(x+3)/(2x+1)=log(1/4)(x-1)/(x-3)
log4(x+3)/(2x+1)=log4(x-3)/(x-1)
(x+3)/(2x+1)=(x-3)/(x-1)
(2x+2)(x-3)=(x+3)(x-1)
2x2-6x+2x-6=x2+2x-3
x2-6x-3=0
x=3±2√3
2011-01-09
展开全部
x=3±2√3
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