Решите 1,2,3 хотябы!

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

$\frac{3(3x-1)-4(2x+3)}{12} =1$

$\frac{9x-3-8x-12}{12}=1$

$\frac{x-15}{12} =1$

$\frac{x-15-12}{12} =0$

$\frac{x-27}{12}=0$

x=27

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

   $\frac{1}{2}x- \frac{1}{3}x=1+ \frac{3}{12}$

   $\frac{1}{6}x= \frac{15}{12}$

   $x= \frac{15}{2}=7,5$

3) $\frac{1}{3}x-1= \frac{1}{4(x+ \frac{4}{5}) }$

   $\frac{1}{3}x-1= \frac{1}{4}x+ \frac{4}{20}$

   $\frac{1}{3}x- \frac{1}{4}x=1+ \frac{1}{5}$

   $\frac{1}{12}x= \frac{6}{5}$

   $x= \frac{72}{5}=14,4$