Cần gấp vote 5 sao Giải pt sau 1/x(x+1) + 1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) = 1/35 (2x-3)(10x+1) < 5x(4x-2)

a) Ta có

$\dfrac{1}{x(x+1)} + \dfrac{1}{(x+1)(x+2)} + \dfrac{1}{(x+2)(x+3)} + \dfrac{1}{(x+3)(x+4)} = \dfrac{1}{35}$

$<-> \dfrac{1}{x} – \dfrac{1}{x+1} + \dfrac{1}{x+1} – \dfrac{1}{x+2} + \dfrac{1}{x+2} – \dfrac{1}{x+3} + \dfrac{1}{x+3} – \dfrac{1}{x+4} = \dfrac{1}{35}$

$<-> \dfrac{1}{x} – \dfrac{1}{x+4} = \dfrac{1}{35}$

$<-> x + 4 – x = \dfrac{1}{35}x(x+4)$

$<-> 140 = x(x+4)$

$<-> x^2 + 4x – 140 = 0$

$<-> (x+14)(x-10) = 0$

$<-> x = 10$ hoặc $x = -14$

Vậy tập nghiệm $S = \{10, -14\}$.

b) Ta có

$(2x-3)(10x + 1) < 5x(4x-2)$

$<-> 20x^2 + 2x – 30x -3 < 20x^2 – 20x$

$<-> 2x – 30x + 20x < 3$

$<-> -8x < 3$

$<-> x > -\dfrac{3}{8}$

Vậy $x > -\dfrac{3}{8}$.