Question

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

\(\frac{x-1}{x}-\frac{3x}{2x-2}=-\frac{5}{2}\)

\(\frac{x+2}{2x-3}-\frac{1}{2x+3}=1-\frac{2x^2-x-4}{4x^2-9}\)

Answer 1

a/ ĐKXĐ: ...

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

\(\Leftrightarrow\frac{x+7}{x^2-1}=\frac{1}{2}\Leftrightarrow2x+14=x^2-1\)

\(\Leftrightarrow x^2-2x-15=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

b/ ĐKXĐ: ...

Đặt \(\frac{x-1}{x}=a\)

\(a-\frac{3}{2a}=-\frac{5}{2}\Leftrightarrow2a^2+5a-3=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-3\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x-1}{x}=-3\\\frac{x-1}{x}=\frac{1}{2}\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x+3\right)-\left(2x-3\right)}{4x^2-9}=\frac{4x^2-9-\left(2x^2-x-4\right)}{4x^2-9}\)

\(\Leftrightarrow2x^2+5x+9=2x^2+x-5\)

\(\Leftrightarrow4x=-14\Rightarrow x=-\frac{7}{2}\)