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

\(A=\dfrac{3x^2-9x+x-3+2}{x-3}\)

\(B=\dfrac{x^2\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x^2+5}{x+2}=x-2+\dfrac{9}{x+2}\)

Để A và B cùng là số nguyên thì

\(\left\{{}\begin{matrix}x-3\in\left\{1;-1;2;-2\right\}\\x+2\in\left\{1;-1;3;-3;9;-9\right\}\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\in\left\{4;2;5;1\right\}\\x\in\left\{-1;-3;1;-5;7;-11\right\}\end{matrix}\right.\)

hay x=1

a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{3x^2-3x+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x\left(x-1\right)+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{\left(x-1\right)\left(3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x+1}{x-2}\)

b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)

Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)

\(\text{x}=1\) (loại)

Vậy giá trị nguyên tập hợp x là:

x=-5;3;9

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

Đk: x \(\ge\)0; x \(\ne\)1; x \(\ne\)9

1) \(B=\left(\frac{2x+3}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)

\(B=\frac{2x+3-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)

\(B=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)

\(B=\frac{-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(B=\frac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+2}{3-\sqrt{x}}\)

2. \(B=\frac{\sqrt{x}+2}{3-\sqrt{x}}=\frac{-\left(3-\sqrt{x}\right)+5}{3-\sqrt{x}}=-1+\frac{5}{3-\sqrt{x}}\)

Để B \(\in\)Z <=> 5 \(⋮\)\(3-\sqrt{x}\)

<=> \(3-\sqrt{x}\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Do \(3-\sqrt{x}\le\)3 => 3 - \(\sqrt{x}\)\(\in\){1; -1; -5}

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