Ответ:

[tex]1)\ \frac{4m+3n-12}{12}[/tex]

[tex]2)\ \frac{5cd-d+c}{cd}[/tex]

Объяснение:

[tex]1)\\\\\frac{m}{3}+\frac{n}{4}-1=\frac{m\cdot 4}{3\cdot 4}+\frac{n\cdot 3}{4\cdot 3}-\frac{1\cdot 3\cdot 4}{3\cdot 4}=\frac{4m}{12}+\frac{3n}{12}-\frac{12}{12}=\frac{4m+3n-12}{12}[/tex]

[tex]2)\\\\5-\frac{1}{c}+\frac{1}{d}=\frac{5\cdot c \cdot d}{ c \cdot d}}-\frac{1\cdot d}{ c \cdot d}}+\frac{1\cdot c}{d\cdot c}=\frac{5cd}{ cd}-\frac{d}{ c}+\frac{ c}{cd}=\frac{5cd-d+c}{cd}[/tex]