Mọi người giúp mình với

Đáp án + Giải thích các bước giải:

$\textbf{B} = \bigg(\dfrac{x +2}{3x} + \dfrac{2}{x + 1} - 3\bigg) : \dfrac{2 - 4x}{x + 1} + \dfrac{x^2 - 3x - 1}{3x}$

ĐKXĐ: $\begin {cases} 3x \ne 0 \\ x +1 \ne 0 \\ 2 - 4x \ne 0 \end {cases}$ hay $x \ne -1; 0; \dfrac{1}{2}$

$\textbf{B} = \bigg(\dfrac{(x + 2)(x + 1)}{3x(x + 1)} + \dfrac{6x}{3x(x + 1)} - \dfrac{9x(x + 1)}{3x(x + 1)}\bigg) \cdot \dfrac{x + 1}{2 - 4x} + \dfrac{x^2 - 3x - 1}{3x}$

$= \bigg(\dfrac{x^2 + 3x + 2}{3x(x + 1)} + \dfrac{6x}{3x(x + 1)} - \dfrac{9x^2 + 9x}{3x(x + 1)}\bigg) \cdot \dfrac{x + 1}{2 - 4x} + \dfrac{x^2 - 3x - 1}{3x}$

$= \dfrac{x^2 + 3x + 2 + 6x - 9x^2 - 9x}{3x(x + 1)} \cdot \dfrac{x + 1}{2 - 4x} + \dfrac{x^2 - 3x - 1}{3x}$

$= \dfrac{-8x^2 + 2}{3x(x + 1)} \cdot \dfrac{x + 1}{2 - 4x} + \dfrac{x^2 - 3x - 1}{3x}$

$= \dfrac{-8x^2 + 2}{3x(2 - 4x)} + \dfrac{x^2 - 3x - 1}{3x}$

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

$= \dfrac{-2(2x + 1)(2x - 1)}{-2(2x - 1)(3x)} + \dfrac{x^2 - 3x -1}{3x}$

$= \dfrac{2x + 1}{3x} + \dfrac{x^2 - 3x - 1}{3x}$

$= \dfrac{x^2 - 3x - 1 + 2x + 1}{3x}$

$= \dfrac{x^2 - x}{3x}$

$= \dfrac{x - 1}{3}$

$\textbf{C} = \bigg(\dfrac{1 + 2x}{4 + 2x} - \dfrac{x}{3x - 6} - \dfrac{2x^2}{3x^2 - 12}\bigg) : \dfrac{24 - 12x}{6 + 13x}$

ĐKXĐ: $\begin {cases} 4 + 2x \ne 0 \\ 3x - 6 \ne 0 \\ 3x^2 - 12 \ne 0 \\ 24 - 12x \ne 0 \\ 6 + 13x \ne 0 \end {cases}$ hay $x \ne -\dfrac{13}{6}; -2; 2$

$\textbf{C} = \bigg(\dfrac{2x + 1}{2(x + 2)} - \dfrac{x}{3(x - 2)} - \dfrac{2x^2}{3(x^2 - 4)}\bigg) \cdot \dfrac{6 + 13x}{12(2 -x)}$

$= \bigg(\dfrac{2x + 1}{2(x + 2)} - \dfrac{x}{3(x - 2)} - \dfrac{2x^2}{3(x + 2)(x - 2)}\bigg) \cdot \dfrac{6 + 13x}{-12(x - 2)}$

$= \bigg(\dfrac{3(2x + 1)(x - 2)}{6(x + 2)(x- 2)} - \dfrac{2x(x + 2)}{6(x + 2)(x - 2)} - \dfrac{4x^2}{6(x + 2)(x - 2)}\bigg) \cdot \dfrac{6 + 13x}{-12(x - 2)}$

$= \bigg(\dfrac{3(2x^2 - 3x - 2)}{6(x + 2)(x - 2)} - \dfrac{2x^2 + 4x}{6(x + 2)(x - 2)} - \dfrac{4x^2}{6(x + 2)(x - 2)}\bigg) \cdot \dfrac{6 + 13x}{-12(x - 2)}$

$= \dfrac{6x^2 - 9x - 6 - 2x^2 - 4x - 4x^2}{6(x + 2)(x - 2)} \cdot \dfrac{6 + 13x}{-12(x - 2)}$

$= \dfrac{-6 - 13x}{6(x + 2)(x - 2)} \cdot \dfrac{6 + 13x}{-12(x - 2)}$

$= \dfrac{(-6 - 13x)(6 + 13x)}{-72(x + 2)(x - 2)^2}$

$= \dfrac{(6 + 13x)^2}{72(x + 2)(x - 2)^2}$