解不等式 1+2log3(x+1)<log3(8x+4)
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前提条件:x>-1/2;x>-1
先将原式化为:log3(3)+2log3(x+1)<log3(8x+4)
log3(3x(x+1)^2)<log3(8x+4)
=>3(x+1)^2<8x+4
解得:-1/3<x<1
先将原式化为:log3(3)+2log3(x+1)<log3(8x+4)
log3(3x(x+1)^2)<log3(8x+4)
=>3(x+1)^2<8x+4
解得:-1/3<x<1
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x>-1
3(x+1)^2<8x+4
得到x属于(-1/3,1)
故解为x属于(-1/3,1)
3(x+1)^2<8x+4
得到x属于(-1/3,1)
故解为x属于(-1/3,1)
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1+2log3(x+1)<log3(8x+4)
=> log3(8x+4) - log3(x+1)^2 > 1 and x+1>0 and 8x+4>0
=> log3[(8x+4)/(x+1)^2] > 1 and x > -1 and x > -1/2
=> (8x+4)/(x+1)^2 > 3 and x > -1/2
=> 3(x+1)^2- (8x+4)/ (x+1)^2< 0 and x > -1/2
=> (3x^2+6x+3-8x-4)/(x+1)^2 > 0 and x > -1/2
=> (3x^2-2x-1)/(x+1)^2<0 and x > -1/2
=> (3x+1)(x-1)/(x+1)^2< 0 and x > -1/2
=> (x< -1 or -1/3 < x<1) and x> -1/2
=> -1/3 < x<1
=> log3(8x+4) - log3(x+1)^2 > 1 and x+1>0 and 8x+4>0
=> log3[(8x+4)/(x+1)^2] > 1 and x > -1 and x > -1/2
=> (8x+4)/(x+1)^2 > 3 and x > -1/2
=> 3(x+1)^2- (8x+4)/ (x+1)^2< 0 and x > -1/2
=> (3x^2+6x+3-8x-4)/(x+1)^2 > 0 and x > -1/2
=> (3x^2-2x-1)/(x+1)^2<0 and x > -1/2
=> (3x+1)(x-1)/(x+1)^2< 0 and x > -1/2
=> (x< -1 or -1/3 < x<1) and x> -1/2
=> -1/3 < x<1
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