Упр.484 ГДЗ Муравин 7 класс (Алгебра)

1) 3/25 + 7/15 и 1 1/18 — 2/27; 3) 1/2 · 3/4 · 5/6 и 1/2 + 3/4 — 5/6;
2) 7/24 + 7/32 и 1 11/45 — 25/36; 4) 3/2 · 5/4 · 7/6 и 3/2 + 5/4 + 7/6.

1. $$\frac{3}{25}+\frac{7}{15}=\frac{9}{75}+\frac{35}{75}=\frac{44}{75}$$
   $$1\frac{1}{18}-\frac{2}{27}=1\frac{3}{54}-\frac{4}{54}=\frac{57}{54}-\frac{4}{54}=\frac{53}{54}$$
   Так как $$\frac{44}{75}<\frac{53}{54},$$ то $$\frac{3}{25}+\frac{7}{15}<1\frac{1}{18}-\frac{2}{27}.$$
2. $$\frac{7}{24}+\frac{7}{32}=\frac{28}{96}+\frac{21}{96}=\frac{49}{96}=\frac{245}{480}$$
   $$1\frac{11}{45}-\frac{25}{36}=1\frac{44}{180}-\frac{125}{180}=\frac{224}{180}=\frac{99}{180}=\frac{264}{480}$$
   Так как $$\frac{245}{480}<\frac{264}{480},$$ то $$\frac{7}{24}+\frac{7}{32}<1\frac{11}{45}-\frac{25}{36}.$$
3. $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}=\frac{3\cdot 5}{2\cdot 4\cdot 6}=\frac{15}{48}=\frac{5}{16}$$
   $$\frac{1}{2}+\frac{3}{4}-\frac{5}{6}=\frac{6}{12}+\frac{9}{12}-\frac{10}{12}=\frac{5}{12}$$
   Так как $$\frac{5}{16}<\frac{5}{12},$$ то $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}<\frac{1}{2}+\frac{3}{4}-\frac{5}{6}.$$
4. $$\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}=\frac{3\cdot 5\cdot 7}{2\cdot 4\cdot 6}=\frac{105}{48}=\frac{35}{16}=2\frac{3}{16}$$
   $$\frac{3}{2}+\frac{5}{4}+\frac{7}{6}=\frac{18}{12}+\frac{15}{12}+\frac{14}{12}=\frac{47}{12}=3\frac{11}{12}$$
   Так как $$2\frac{3}{16}<3\frac{11}{12},$$ то $$\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}<\frac{3}{2}+\frac{5}{4}+\frac{7}{6}.$$

Ответ

1) $$\frac{3}{25}+\frac{7}{15}<1\frac{1}{18}-\frac{2}{27}$$
2) $$\frac{7}{24}+\frac{7}{32}<1\frac{11}{45}-\frac{25}{36}$$
3) $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}<\frac{1}{2}+\frac{3}{4}-\frac{5}{6}$$
4) $$\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}<\frac{3}{2}+\frac{5}{4}+\frac{7}{6}$$