=2x+\(\frac{1}{x}\)-5\(x^4\)

b) \(\frac{4\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)\(-\frac{7x+3}{\left(2x-1\right)\left(2x+1\right)}\)\(=\frac{8x+1-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\frac{x-2}{\left(2x-1\right)\left(2x+1\right)}\)

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

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