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(вложение)

 

     

     

     

       

     

    1/(1-x)\1. (1-x > 0, x < 1)\ a. (1-x)/2x-1 > 0 (2x > 0)\ (1-x)-2x > 0\ 1-3x > 0\ -3x > -1\ x < 1/3\ b. (2x < 0)\ (1-x)/2x-1 < 0\ 1-3x < 0\ -3x < -1\ x > 13\ 2.   (1-x < 0, x > 1)\ (1-x)/2x-1 < 0\ a. (2x >0)\(1-x)-2x < 0\ 1-3x < 0\ -3x < -1\ x > 1/3\ b. (2x < 0)\ (1-x)-2x > 0\ 1-3x > 0\ -3x > -1\ x < 1/3\\ (01)' alt='5) 1/2x > 1/(1-x)\1. (1-x > 0, x < 1)\ a. (1-x)/2x-1 > 0 (2x > 0)\ (1-x)-2x > 0\ 1-3x > 0\ -3x > -1\ x < 1/3\ b. (2x < 0)\ (1-x)/2x-1 < 0\ 1-3x < 0\ -3x < -1\ x > 13\ 2.   (1-x < 0, x > 1)\ (1-x)/2x-1 < 0\ a. (2x >0)\(1-x)-2x < 0\ 1-3x < 0\ -3x < -1\ x > 1/3\ b. (2x < 0)\ (1-x)-2x > 0\ 1-3x > 0\ -3x > -1\ x < 1/3\\ (01)' align='absmiddle' class='latex-formula'>

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    = -2\ 1. (x-1>0, x>1)\ 1 >= -2(x-1)\ -1/2 =< (x-1)\ 1/2 =< x\2. (x-1<0, x<1)\ 1 =< -2(x-1) \ -1/2 >= (x-1)\ 1/2 >= x\ (x =< 1/2) and (x>1)\\4) x -3 + 4/(x+1) > 0\ 1. ((x+1)>0, x>-1)\4/(x+1) > 3 - x\ 4 > (3-x)(x+1)\ 4 > 3x + 3 -x^2 -x\ 0 > -x^2 +2x -1\ 0 < x^2-2x+1\ 0 < (x-1)^2, x <> 1 \2. ((x+1)<0, x<-1)\ 4/(x+1) < 3 - x\ 4 < (3-x)(x+1)\ 4 < 3x + 3 -x^2 -x\ 0 < -x^2 +2x -1\ 0> x^2-2x+1\ 0 > (x-1)^2' alt='3) 1/(x-1) >= -2\ 1. (x-1>0, x>1)\ 1 >= -2(x-1)\ -1/2 =< (x-1)\ 1/2 =< x\2. (x-1<0, x<1)\ 1 =< -2(x-1) \ -1/2 >= (x-1)\ 1/2 >= x\ (x =< 1/2) and (x>1)\\4) x -3 + 4/(x+1) > 0\ 1. ((x+1)>0, x>-1)\4/(x+1) > 3 - x\ 4 > (3-x)(x+1)\ 4 > 3x + 3 -x^2 -x\ 0 > -x^2 +2x -1\ 0 < x^2-2x+1\ 0 < (x-1)^2, x <> 1 \2. ((x+1)<0, x<-1)\ 4/(x+1) < 3 - x\ 4 < (3-x)(x+1)\ 4 < 3x + 3 -x^2 -x\ 0 < -x^2 +2x -1\ 0> x^2-2x+1\ 0 > (x-1)^2' align='absmiddle' class='latex-formula'>

     

    1)' alt='(-1 < x < 1) and (x > 1)' align='absmiddle' class='latex-formula'>

     

    не при каких x. Решение: