a,\(2x\left(x-3\right)=x-3.\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy .....
b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow x^2+2x-5x+10=8\)
\(\Leftrightarrow x^2-3x+10-8=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy ....
\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12.\)
\(\Leftrightarrow\frac{\left(2x+1\right)\cdot3}{4\cdot3}-\frac{\left(x-5\right)\cdot4}{3\cdot4}< \frac{4x-1}{12}+12.\)
\(\Leftrightarrow\frac{6x+3}{12}-\frac{4x-20}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{6x+3-4x+20}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{2x+23}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{2x+23-4x+1}{12}< 12\)
\(\Leftrightarrow\frac{-2x+24}{12}< 12\)
\(\Leftrightarrow-2x+24< 144\)
\(\Leftrightarrow-2x< 120\)
\(\Leftrightarrow x< -60\)
