Упр.5 ГДЗ Мерзляк Полонский 7 класс (Алгебра)

1) 2,6 и 2,8; 4) 9/11 и 17/22; 7) -19 и 0;
2) 15,7 и 15,69; 5) 7/12 и 11/18; 8) 93 и -95;
3) 0,03 и 0,08; 6) 7/8 и 8/9; 9) -0,23 и -0,3. Вычислите значение числового выражения:
1) 14*7/15 — 3*3/23*23/27-1*1/5*1/6;
2) (5*8/9:1*17/36+1*1/4) * 5/21;
3) (-3,25 — 2,75) : (-0,6) + 0,8 * (-7);
4) (-1*3/8 — 2*5/12): 5*5/12.

1. Сравним числа:

   $$2,6<2,8$$

   $$15,7>15,69$$

   $$0,03<0,08$$

   $$\frac{9}{11}=\frac{18}{22},\quad \frac{18}{22}>\frac{17}{22}$$

   $$\frac{7}{12}=\frac{21}{36},\quad \frac{21}{36}<\frac{22}{36}$$

   $$\frac{7}{8}=\frac{63}{72},\quad \frac{63}{72}<\frac{64}{72}$$

   $$-19<0$$

   $$93>-95$$

   $$-0,23>-0,3$$

2. Вычислим значение выражения:

   $$14\frac{7}{15}-3\frac{3}{23}\cdot\frac{23}{27}-1\frac{1}{5}\cdot\frac{1}{6}$$

   $$=14\frac{7}{15}-\frac{8}{3}-\frac{1}{5}$$

   $$=14\frac{7}{15}-2\frac{5}{15}-\frac{3}{15}$$

   $$=14\frac{7}{15}-2\frac{8}{15}$$

   $$=11\frac{12}{15}=11\frac{4}{5}$$

3. $$\left(5\frac{8}{9}:1\frac{17}{36}+1\frac{1}{4}\right)\cdot\frac{5}{21}$$

   $$=\left(\frac{53}{9}:\frac{53}{36}+\frac{5}{4}\right)\cdot\frac{5}{21}$$

   $$=\left(\frac{53}{9}\cdot\frac{36}{53}+\frac{5}{4}\right)\cdot\frac{5}{21}$$

   $$=\left(4+\frac{5}{4}\right)\cdot\frac{5}{21}$$

   $$=5\frac{1}{4}\cdot\frac{5}{21}=\frac{21}{4}\cdot\frac{5}{21}=\frac{5}{4}=1\frac{1}{4}$$

4. $$(-3,25-2,75):(-0,6)+0,8\cdot(-7)$$

   $$=-6:(-0,6)-5,6$$

   $$=10-5,6=4,4$$

5. $$\left(-1\frac{3}{8}-2\frac{5}{12}\right):5\frac{5}{12}$$

   $$=\left(-\frac{11}{8}-\frac{29}{12}\right):\frac{65}{12}$$

   $$=\left(-\frac{33}{24}-\frac{58}{24}\right):\frac{65}{12}$$

   $$=-\frac{91}{24}\cdot\frac{12}{65}=-\frac{7}{10}$$

Ответ

1) $$2,6<2,8$$; $$15,7>15,69$$; $$0,03<0,08$$; $$\frac{9}{11}>\frac{17}{22}$$; $$\frac{7}{12}<\frac{11}{18}$$; $$\frac{7}{8}<\frac{8}{9}$$; $$-19<0$$; $$93>-95$$; $$-0,23>-0,3$$.

2) $$11\frac{4}{5}$$.

3) $$1\frac{1}{4}$$.

4) $$4,4$$.

5) $$-\frac{7}{10}$$.