Giải phương trình 5/x+4/x+1=3/x+2+2/x+3

5/x + 4/(x + 1) = 3/(x + 2) + 2/(x + 3)     (ĐK: x \ne 0,x \ne -1,x \ne -2, x \ne – 3)

<=> (5/x + 1) + (4/(x + 1) + 1) = (3/(x + 2) + 1) + (2/(x + 3) + 1)

<=> (x + 5)/x + (4 + x + 1)/(x + 1) = (3 + x + 2)/(x + 2) + (2 + x + 3)/(x + 3)

<=> (x + 5)/x + (x + 5)/(x + 1) = (x + 5)/(x + 2) + (x + 5)/(x+ 3)

<=> (x + 5)/x + (x + 5)/(x + 1) – (x + 5)/(x + 2) – (x + 5)/(x+ 3) = 0

<=> (x + 5)(1/x + 1/(x + 1) – 1/(x + 2) – 1/(x + 3)) = 0

+) x + 5 = 0

<=> x  = -5 (thỏa mãn)

+) 1/x + 1/(x + 1) – 1/(x + 2) – 1/(x + 3) = 0

<=> ((x + 1)(x + 2)(x + 3) + x(x + 2)(x + 3) – x(x + 1)(x + 3) – x(x + 1)(x + 2))/(x(x + 1)(x + 2)(x + 3)) = 0

<=> (x^2 + 3x + 2)(x + 3) + (x^2 + 2x)(x + 3) – (x^2 + 1)(x + 3) – (x^2 + x)(x + 2) = 0

<=> x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 + x^3 + 3x^2 + 2x^2 +6x – x^3 – 3x^2 – x – 3 – x^3 – 2x^2 – x^2 – 2x = 0

<=> 5x^2 + 14x + 3 = 0

<=> 5(x^2 + (14)/5x + 3/5) = 0

<=> 5[x^2 + 2. x (14)/(10) + (14/10)^2 – (34)/(25)] = 0

<=> (x + (14)/(10))^2 – (34)/(25) = 0

<=> (x + (7)/(5) – (\sqrt{34})/5)(x + (7)/(5) + (\sqrt{34})/5) = 0

<=> (x + (7 – \sqrt{34})/5)(x + (7 + \sqrt{34})/5) = 0

<=> x + (7 – \sqrt{34})/5 = 0 hoặc x + (7 + \sqrt{34})/5 = 0

<=> x = (-7 + \sqrt{34})/5 (thỏa mãn)  hoặc x  = (-7 – \sqrt{34})/5 (thỏa mãn)

Vậy S = {-5,(-7 + \sqrt{34})/5,(-7- \sqrt{34})/5} là tập nghiệm của pt