Question

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

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

To solve the given inequality $\frac{1}{(2x+1)(2x+2)} + \frac{1}{(2x+2)(2x+3)} + \frac{1}{(2x+3)(2x+4)} \leq \frac{3}{4}$, we need to find the values of $x$ that satisfy this inequality.

First, find a common denominator for the fractions on the left-hand side of the inequality. The common denominator for the three fractions is $(2x+1)(2x+2)(2x+3)(2x+4)$. Multiplying both sides of the inequality by this common denominator, we get:

$(2x+2)(2x+3) + (2x+1)(2x+3) + (2x+1)(2x+2) \leq \frac{3}{4} \times (2x+1)(2x+2)(2x+3)(2x+4)$

Now, expand and simplify the inequality:

$4x^2 + 10x + 6 + 4x^2 + 8x + 3 + 4x^2 + 6x + 2 \leq \frac{3}{4} \times (2x+1)(2x+2)(2x+3)(2x+4)$

$12x^2 + 24x + 11 \leq \frac{3}{4} \times (2x+1)(2x+2)(2x+3)(2x+4)$

Now, expand the right-hand side and simplify the inequality further:

$12x^2 + 24x + 11 \leq \frac{3}{4} \times (4x^2 + 20x + 24)(4x^2 + 20x + 24)$

$12x^2 + 24x + 11 \leq \frac{3}{4} \times (16x^4 + 160x^3 + 552x^2 + 640x + 240)$

$12x^2 + 24x + 11 \leq 12x^4 + 120x^3 + 414x^2 + 480x + 180$

Now, move all terms to one side of the inequality to set it to zero:

$12x^4 + 120x^3 + 402x^2 + 456x + 169 - 12x^2 - 24x - 11 \geq 0$

$12x^4 + 120x^3 + 390x^2 + 432x + 158 \geq 0$

Unfortunately, this polynomial cannot be factored nicely, and its solutions are not simple expressions. You may need to use numerical methods or a graphing calculator to find approximate solutions for this inequality.