b)
x+3/x+1 + x+1 / x = 2 (ĐKXĐ : x<>0 ; x <>-1)
<=> x(x+3)/x(x+1) + (x+1)(x+1)/x(x+1) = 2x(x+1)/x(x+1)
<=> x(x+3) + (x+1)(x+1) = 2x(x+1)
<=> x^2 + 3x + (x+1)(x+1) - 2x(x+1)=0
<=> x^2 + 3x +(x+1)(x+1-2x)=0
<=> x^2 + 3x + (x+1)(-x+1)=0
<=> x^2 + 3x -x^2 + x - x + 1 = 0
<=> 3x + 1 = 0
<=> 3x = -1
<=> x = -1/3
Vậy tập nghiệm của pt :
S={-1/3}

Bài 2:
a) (x-3)^2 – 12 < (x-1)(x+3)
<=> x^2 - 6x + 9 - 12 < x^2 + 3x - x - 3
<=> x^2 - x^2 - 6x - 3x + x < -3 - 9 + 12
<=> -10x < 0
<=> x > 0
Vậy S = { x/ x > 0}
b) 4x - 5/ - 3 > 7 - x / -5
<=> -5(4x-5)/ 15 > -3(7-x)/ 15
<=> -5(4x-5) > -3(7-x)
<=> -20x + 25 > -21 + 3x
<=> -20x - 3x > - 21 - 25
<=> -23x > - 46
<=> 23x < 46
<=> x < 2
Vậy S = { x / x<2}

Đề 2:
Bài 1:
b) 3x(x-2) = 3(x-2)
<=> 3x(x-2) - 3(x-2) = 0
<=> (3x-3)(x-2) = 0
<=> 3x - 3 = 0 hoặc x - 2 =0
TH1 : 3x - 3 = 0
<=> 3x = 3
<=> x = 1
TH2 : x - 2 = 0
<=> x = 2
Vậy tập nghiệm của pt :
S={ 2;1}

d) x+1 / 2x-2 – x -1/2x+2 + 2/1 – x^2 = 0 (ĐKXĐ : X<>+-1)
<=> (x+1)(2x+2)/(2x-2)(2x+2) - (x-1)(2x+2)/(2x+2)(2x-2) - 2 / (x+2)(x-2) =0
=> (x+1)(2x+2) - (x-1)(2x+2) - 2 = 0
<=>(2x+2)(x+1-x+1)-2=0
<=> 2.(2x+2) - 2 = 0
<=> 4x + 4 - 2 =0
<=> 4x + 2 = 0
<=> 4x = -2
<=> x = -2/4 = -1/2 (Thỏa ĐK)
Vậy S = { -1/2}

Bài 2:
a) 2x+1/3 – 3x -2 / 2 <= 3
<=> 2(2x+1)/6 - 3(3x-2)/6 <= 18/6
<=> 2(2x+1) - 3(3x-2) <=18
<=> 4x + 2 - 9x + 6 <=18
<=> -5x + 8 <=18
<=.> -5x <= 18-8
<=> -5x <=10
<=> 5x >= -10
<=> x >=-2
Vậy S = { x/x >=-2}

Bài 1:
a. (4x-5)(x+3)-(2x-3)(7+2x)=0
<---> 4x²+12x-5x-15-(14x+4x²-2x-6x)=0
<---> 4x²+12x-5x-15-(8x+4x²-21)=0
<---> 4x²+12x-5x-15-8x-4x²+21=0
<---> 12x-5x-15-8x+21=0
<---> -x+6=0
<---> -x=-6
<---> x=6
S={6}

b. x+3/x+1+x-1/x=2
<---> x(x+3)/x(x+1)+(x-1)(x+1)/x(x+1)=2x(x+1)/x(x+1)
<---> x²+3x/x²+x+x²+x-x-1/x²+x=2x²+2x/x²+x
<---> x²+3x+x²+x-x-1=2x²+2x
<---> 2x²+3x-1=2x²+2x
<---> 2x²+3x-1-2x²-2x=0
<---> x-1=0
<---> x=1
S={1}