a,[tex]\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x-\frac{x-3}{2}}{5}-x+1[/tex]
b, [tex]\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}[/tex]
c,[tex]\frac{5}{x-1}-\frac{4}{3-6x+3x^2}=3[/tex]
d, [tex]\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10[/tex]
d.
$ \frac{x - 241}{17} + \frac{x - 220}{19} + \frac{x - 195}{21} + \frac{x - 166}{23} = 10 \\ \Leftrightarrow \frac{x - 241}{17} - 1 + \frac{x - 220}{19} - 2 + \frac{x - 195}{21} - 3 + \frac{x - 166}{23} - 4 = 0 \\ \Leftrightarrow \frac{x - 258}{17} + \frac{x - 258}{19} + \frac{x - 258}{21} + \frac{x - 258}{23} = 0 \\ ... $
c.
$ \frac{5}{x - 1} - \frac{4}{3 - 6x + 3x^2} = 3 \\ \Leftrightarrow \frac{5}{x - 1} - \frac{4}{3(x^2 - 2x + 1)} - 3 = 0 \\ \Leftrightarrow \frac{15x - 15}{3(x - 1)^2} - \frac{4}{3(x - 1)^2} - \frac{9x^2 - 18x + 9}{3(x - 1)^2} = 0 \\ \Leftrightarrow \frac{-9x^2 + 33x - 28}{3(x - 1)^2} = 0 \\\Rightarrow -9x^2 + 33x - 28 = 0 \\ \Leftrightarrow -9x^2 + 12x + 21x - 28 = 0 \\ \Leftrightarrow (-9x + 12)\left (x - \frac{7}{3} \right ) = 0 \\ \Leftrightarrow -9\left (x - \frac{4}{3} \right ) \left ( x - \frac{7}{3} \right ) = 0 \\ \Leftrightarrow ... $
b.
$ \frac{2 - x}{2001} - 1 = \frac{1 - x}{2002} - \frac{x}{2003} \\ \Leftrightarrow \frac{2 - x}{2001} +1 - 1 + 1 = \frac{1 - x}{2002} + 1 - \frac{x}{2003} + 1 \\ \Leftrightarrow \frac{2003 - x}{2001} = \frac{2003 - x}{2002} + \frac{2003 - x}{2003} \\ \Leftrightarrow \frac{2003 - x}{2001} - \frac{2003 - x}{2002} - \frac{2003 - x}{2003} = 0 \\ \Leftrightarrow ... $