\(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+4x}{4-x^2}\)

a) \(=\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{x^2+4x}{x^2-4}\)

\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2-3x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+3x+2+x^2-3x+2-x^2-4x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

b) Tại x = 4 thỏa mãn x ≠ ±2 => A = 1/3

c) Ta có \(A=\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=1-\frac{4}{x+2}\)

Để A nhận giá trị nguyên dương thì \(\frac{4}{x+2}\)nguyên

=> 4 chia hết cho x+2

=> x + 2 thuộc Ư(4) = { ±1 ; ±2 ; ±4 }

Rồi bạn thử từng cái vô xem cái nào làm cho A dương thì lấy

        Bài làm :

1 m/s = 3,6 km/h

=> 15m/s = 15.3,6 = 54 km/h

Vậy 15m/s = 54 km/h

( 6x³ + 15x² - 4x - 10 ) : ( 3x² - 2 )

= [ ( 6x³ - 4x ) + ( 15x² - 10 ) ] : ( 3x² - 2 )

= [ 2x( 3x² - 2 ) + 5( 3x² - 2 ) ] : ( 3x² - 2 )

= ( 3x² - 2 )( 2x + 5 ) : ( 3x² - 2 )

= 2x + 5 

\(A=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

a) ĐKXĐ : x ≠ ±2

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

b) Để A = 1 => \(\frac{1}{x-2}=1\)=> x - 2 = 1 => x = 3 ( tm )

c) Để A > 1 => \(\frac{1}{x-2}>1\)

=> \(\frac{1}{x-2}-1>0\)

=> \(\frac{1}{x-2}-\frac{x-2}{x-2}>0\)

=> \(\frac{1-x+2}{x-2}>0\)

=> \(\frac{-x+3}{x-2}>0\)

Xét hai trường hợp

1. \(\hept{\begin{cases}-x+3>0\\x-2>0\end{cases}}\Rightarrow\hept{\begin{cases}-x>-3\\x>2\end{cases}}\Rightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}}\Rightarrow2< x< 3\)

2. \(\hept{\begin{cases}-x+3< 0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}-x< -3\\x< 2\end{cases}}\Rightarrow\hept{\begin{cases}x>3\\x< 2\end{cases}}\)( loại )

Vậy với 2 < x < 3 thì A > 1

d) Để A nguyên => \(\frac{1}{x-2}\)nguyên

=> 1 ⋮ x - 2

=> x - 2 ∈ Ư(1) = { ±1 }

=> x ∈ { 1 ; 3 } thì A nguyên

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

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

\(=\frac{2x}{1-x}\)