1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
               \(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
               \(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
                          \(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
                          \(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
                               \(2x=\frac{53}{30}\)
                                  \(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
                               \(2x=\frac{37}{30}\)
                                  \(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
    \(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
                          \(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
         \(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
            \(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
                 \(-\frac{5}{7}x=-\frac{11}{45}\)
                           \(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}

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

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

\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

\(\Rightarrow x=\frac{16}{11}\)

\(\Leftrightarrow\frac{3x}{6}+\frac{5}{6}=\frac{-4}{6}\)

\(\Leftrightarrow3x+5=-4\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\)

\(\frac{x}{2}+\frac{5}{6}=\frac{-2}{3}\)

\(\Rightarrow\frac{x}{2}=\frac{-2}{3}-\frac{5}{6}=\frac{-4}{6}-\frac{5}{6}=\frac{-4-5}{6}=\frac{-9}{6}=\frac{-3}{2}\)

\(\Rightarrow\frac{x}{2}=\frac{-3}{2}\)

\(\Rightarrow x=-3\)

x/2 + 5/6 = -2/3

         x/2 = -2/3 - 5/6

         x/2 = -3/2

=> x = -3

ĐK của A \(x\ne4\),ĐK của B \(\hept{\begin{cases}x\ne0\\x\ne5\end{cases}}\)

a, \(x^2-3x=0\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Với \(x=0\Rightarrow A=\frac{-5}{-4}=\frac{5}{4}\)

Với \(x=3\Rightarrow A=\frac{3-5}{3-4}=2\)

b. \(B=\frac{x+5}{2x}+\frac{x-6}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}=\frac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\frac{x^2-10x+25}{2x\left(x-5\right)}=\frac{\left(x-5\right)^2}{2x\left(x-5\right)}=\frac{x-5}{2x}\)

c. \(P=\frac{A}{B}=\frac{x-5}{x-4}.\frac{2x}{x-5}=\frac{2x}{x-4}=\frac{2x-8}{x-4}+\frac{8}{x-4}=2+\frac{8}{x-4}\)

P nguyên \(\Leftrightarrow x-4\inƯ\left(8\right)\Rightarrow x-4\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow x\in\left\{-4;0;2;3;5;6;8;12\right\}\)

So sánh điều kiện ta thấy \(x\in\left\{-4;2;3;6;8;12\right\}\)thì P nguyên