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) 24/11/2021 Bởi Rylee 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 $x > -\dfrac{3}{8}$. Bình luận
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 $x > -\dfrac{3}{8}$.
Trong hình vẽ !