Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = 2\\
x =  - 1\\
x = 1\\
x = 0
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = 2\\
x =  - 2\\
x = 0
\end{array} \right.\\
g)\left[ \begin{array}{l}
x = 12\\
x =  - 2\\
x = 6\\
x = 4
\end{array} \right.\\
h)\left[ \begin{array}{l}
x = 5\\
x =  - 3\\
x = 3\\
x =  - 1\\
x = 2\\
x = 0
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \dfrac{1}{2}\\
\dfrac{3}{{2x - 1}} \in Z \to 2x - 1 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 =  - 3\\
2x - 1 = 1\\
2x - 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x =  - 1\\
x = 1\\
x = 0
\end{array} \right.\\
b)\dfrac{5}{{{x^2} + 1}} \in Z \to {x^2} + 1 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
{x^2} + 1 = 5\\
{x^2} + 1 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} = 4\\
{x^2} = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x =  - 2\\
x = 0
\end{array} \right.\\
g)DK:x \ne 5\\
\dfrac{{x + 2}}{{x - 5}} = \dfrac{{x - 5 + 7}}{{x - 5}} = 1 + \dfrac{7}{{x - 5}}\\
\dfrac{{x + 2}}{{x - 5}} \in Z \to \dfrac{7}{{x - 5}} \in Z\\
 \to x - 5 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
x - 5 = 7\\
x - 5 =  - 7\\
x - 5 = 1\\
x - 5 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x =  - 2\\
x = 6\\
x = 4
\end{array} \right.\\
h)DK:x \ne 1\\
\dfrac{{{x^2} + 3}}{{x - 1}} = \dfrac{{{x^2} - 1 + 4}}{{x - 1}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 4}}{{x - 1}}\\
 = \left( {x + 1} \right) + \dfrac{4}{{x - 1}}\\
\dfrac{{{x^2} + 3}}{{x - 1}} \in Z \to \dfrac{4}{{x - 1}} \in Z\\
 \to x - 1 \in U\left( 4 \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 4\\
x - 1 =  - 4\\
x - 1 = 2\\
x - 1 =  - 2\\
x - 1 = 1\\
x - 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x =  - 3\\
x = 3\\
x =  - 1\\
x = 2\\
x = 0
\end{array} \right.
\end{array}\)

a) 2x-1=3 ⇒x=4;b) $x^{2}$ +1=5⇒x=2;c) $x^{2}$ -x+1=7⇒x=-2

d) $x^{2}$ -59=$x^{2}$ -64+5⇒x+8∈Ư(5)⇒x∈(-9;-7;-3;-13)

e)x+2=0 thì phân thức sẽ có giá trị nguyên⇒x=-2;

g)(tương tự phần d)x∈(-2;12;6;4)

h)x=-1