câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

          \(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);

          \(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\) 

         \(\dfrac{1}{x^2-1}\)  =  \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

\(\dfrac{x}{2x-4}\)  =   \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\) 

  \(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

Lời giải:
a. ĐKXĐ: \(\left\{\begin{matrix} 3x\neq 0\\ x+1\neq 0\\ 2-4x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\neq -1\\ x\neq \frac{1}{2}\end{matrix}\right.\)

b.

\(P=\left[\frac{(x+2)(x+1)+3x.2}{3x(x+1)}-3\right].\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x(x+1)}{3x(x+1)}.\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-2(2x-1)(2x+1)(x+1)}{6x(x+1)(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

Tại $x=2023$ thì:

$P=\frac{2023-1}{3}=\frac{2022}{3}=674$

c.

Để $P$ nguyên thì $x-1\vdots 3$
$\Rightarrow x=3k+1$ với $k$ nguyên bất kỳ. 

Kết hợp với ĐKXĐ thì $x=3k+1$ với $k\in\mathbb{Z}$

1 How much time do you spend playing video gé on weekends

2 Which family owns the largest field for planting crops in the rural area

1 How much time do you spendplaying video games on weekends?

2 Which family owns the largest field for planting crops in the rural area?

1 I enjoy going to the park for a picnic for it allows me to relax in nature

2 We should go to this village in order to explore traditional ways of life

3 If we throw trash on the ground, the environment gets dirty

1 There are too many cars in some cities, and this causes air pollution

2 She is crazy about taking photographs of beautiful landscapes and scenery 

3 The river water is dirty enough to pose threat to aquatic life

1 There are too many cars in some cities, and this causes air pollution.

2 She is crazy about taking photographs of beautiful landscapes and scenery.

3 The river water is dirty enough to pose threat to aquatic life.

1 Avoid touching hot surfaces as they can burn you

2 Are bicycles a popular mode of transport in the countryside

\(B=\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\left(x\ne0;y\ne\pm2x\right)\)

\(=\dfrac{2x+y}{x\left(2x-y\right)}-\dfrac{8y}{4x^2-y^2}+\dfrac{2x-y}{x\left(2x+y\right)}\)

\(=\dfrac{\left(2x+y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}-\dfrac{8xy}{x\left(2x-y\right)\left(2x+y\right)}+\dfrac{\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{8x^2-8xy+2y^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{2\left(4x^2-4xy+y^2\right)}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{2\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{2\left(2x-y\right)}{x\left(2x+y\right)}\)

\(=\dfrac{4x-2y}{2x^2+xy}\)

Với \(x\ne0;y\ne\pm2x\), xét: \(x=\dfrac{1}{2};y=-\dfrac{3}{2}\left(tmdk\right)\)

Thay \(x=\dfrac{1}{2};y=-\dfrac{3}{2}\) vào \(B\), ta được:

\(B=\dfrac{4\cdot\dfrac{1}{2}-2\cdot\dfrac{-3}{2}}{2\cdot\left(\dfrac{1}{2}\right)^2+\dfrac{1}{2}\cdot\dfrac{-3}{2}}=\dfrac{5}{-\dfrac{1}{4}}=-20\)

\(Toru\)