Question

ПОМОГИТЕ ПОЖАЛУЙТА!!!!!! НЕ БОЛЬШОЙ НОМЕР ПО МАТЕМАТИКЕ. ВЗОРАНИЕ СПОСИБО

Answer 1

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Answer 2

$\displaystyle \tt 1). \ \ 7\frac{1}{2}=7+\frac{1\cdot9}{2\cdot9}=7\frac{9}{18}\\\\{} \ \ \ \ \ \ 4\frac{2}{3}=4+\frac{2\cdot6}{3\cdot6}=4\frac{12}{18}\\\\{} \ \ \ \ \ \ 5\frac{1}{6}=5+\frac{1\cdot3}{6\cdot3}=5\frac{3}{18}\\\\{} \ \ \ \ \ \ 3\frac{4}{9}=3+\frac{4\cdot2}{9\cdot2}=3\frac{8}{18}$

$\displaystyle \tt 2). \ \ 9\frac{1}{2}=9+\frac{1\cdot6}{2\cdot6}=9\frac{6}{12}\\\\{} \ \ \ \ \ \ 5\frac{1}{3}=5+\frac{1\cdot4}{3\cdot4}=5\frac{4}{12}\\\\{} \ \ \ \ \ \ 6\frac{3}{4}=6+\frac{3\cdot3}{4\cdot3}=6\frac{9}{12}\\\\{} \ \ \ \ \ \ 8\frac{5}{6}=8+\frac{5\cdot2}{6\cdot2}=8 \frac{10}{12}$

$\displaystyle \tt 3). \ \ 12\frac{2}{3}=12+ \frac{2\cdot15}{3\cdot15}=12 \frac{30}{45}\\\\{} \ \ \ \ \ \ \ 7\frac{3}{5}=7+ \frac{3\cdot9}{5\cdot9}=7 \frac{27}{45}\\\\{} \ \ \ \ \ \ \ 6\frac{4}{9}=6+\frac{4\cdot5}{9\cdot5}=6\frac{20}{45}\\\\{} \ \ \ \ \ \ \ 4 \frac{2}{15}=4+ \frac{2\cdot3}{15\cdot3}=4 \frac{6}{45}$