a) ta có: \(A=\frac{2x}{x-2}=\frac{2x-4+4}{x-2}=\frac{2.\left(x-2\right)+4}{x-2}=\frac{2.\left(x-2\right)}{x-2}+\frac{4}{x-2}=2+\frac{4}{x-2}\)

Để \(A\inℤ\)

\(\Rightarrow\frac{4}{x-2}\inℤ\)

\(\Rightarrow4⋮x-2\Rightarrow x-2\inƯ_{\left(4\right)}=\left(4;-4;2;-2;1;-1\right)\)

nếu x -2 = 4 => x = 6 (TM)

x- 2= - 4 => x= - 2 (TM)

x- 2= 2 => x = 4 (TM)

x- 2 = -2 => x = 0 (TM)

x - 2 = 1 => x = 3 (TM) 

x - 2 = -1 => x=  1 (TM)

KL: \(x\in\left(6;-2;4;0;3;1\right)\)

c) ta có: \(C=\frac{x^2+2}{x+1}=\frac{\left(x+1\right).\left(x-1\right)+3}{x+1}=\frac{\left(x+1\right).\left(x-1\right)}{x+1}+\frac{3}{x+1}\)\(=x-1+\frac{3}{x+1}\)

Để \(C\inℤ\)

\(\Rightarrow\frac{3}{x+1}\inℤ\)

\(\Rightarrow3⋮x+1\Rightarrow x+1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)

nếu x + 1 = 3 => x = 2 (TM)

x + 1 = - 3 => x = -4 (TM)

x + 1 = 1 => x = 0 

x + 1 = -1 => x = -2 (TM)

KL: \(x\in\left(2;-4;0;-2\right)\)

p/s

a) Giả sử \(C=\frac{2x+3}{7}=t\left(t\in Z\right)\)

\(\Rightarrow x=\frac{7t-3}{2}\). Để \(x\in Z\) thì t phải lẻ. Nói cách khác \(t=2k+1\left(k\in Z\right)\)

Suy ra  \(x=\frac{7\left(2k+1\right)-3}{2}=14k+2\)

Vậy để \(\frac{2x+3}{7}\in Z\) thì \(x=14k+2\left(k\in Z\right)\)

b) Ta thấy \(C=\frac{6x-1}{3x+2}=\frac{\left(6x+4\right)-5}{3x+2}=2-\frac{5}{3x+2}\)

Do x nguyên nên C đạt GTNN khi \(\frac{5}{3x+2}\) lớn nhất. Điều này xảy ra khi 3x + 2 = 2 hay x = 0.

Vậy \(minC=-\frac{1}{2}\) khi x = 0.

THANKS NHA!

a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)

 \(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)

\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa

\(\frac{5-3x}{2x-1}\in Z\Rightarrow\frac{10-6x}{2x-1}\in Z\)

\(\frac{10-6x}{2x-1}=\frac{10-6x+3-3}{2x-1}=\frac{7-3\left(2x-1\right)}{2x-1}=\frac{7}{2x+1}-3\)

Để \(\frac{7}{2x-1}-3\in Z\Leftrightarrow\frac{7}{2x-1}\in Z\)

=> 2x - 1 ∈ Ư(7) = { ± 1; ± 7 }

Ta có : 2x - 1 = 7 => 2x = 8 => x = 4

           2x - 1 = 1 => 2x = 2 => x = 1

           2x - 1 = - 1 => 2x = 0 => x = 0

           2x - 1 = - 7 => 2x = - 6 => x = - 3

Vậy x = { - 3; 0; 1; 4 }