решите уровнение 23:(X-1\3)=2 и есчё одно 5\4=(t+1\2)\(1\3)

[tex]\frac{23}{x- \frac{1}{3}} =2 \\ x \neq \frac{1}{3} \\ \frac{23}{x- \frac{1}{3}} =2|*(x- \frac{1}{3})\\ 23=2*(x-\frac{1}{3}) \\ 2x-\frac{2}{3}=23 \\ 2x=23\frac{2}{3} \\ x=\frac{71}{6}=11\frac{5}{6}[/tex]

[tex] \frac{5}{4} = \frac{t+ \frac{1}{2} }{ \frac{1}{3}}|* \frac{1}{3}} \\ \frac{5}{12}=t+ \frac{1}{2}|*12 \\ 5=12t+6 \\ 12t=-1 \\ t=- \frac{1}{12} [/tex]

23:(х-1/3)=2 X-1/3=23/2 X=23/2+1/3 X=71/6 X=11 5/6