Упр.4 ГДЗ Мерзляк Полонский 7 класс (Алгебра)
1) 0,42; 2) 0,072; 3) 4,03; 4) 6,125.
Чему равно значение выражения:
1) 18*5/12 — 7/12 *1*19/21 — 17/12*2/3;
2) (6*3/4 — 5*1/8 :1*9/32) * 5/11;
3) (-1,42 — (-3,22)) : (-0,4) + (-6) * (-0,7);
4) (-7/18+11/22): (-19/48);
5) (-3*1/12- 2*1/15): (-5*3/20)?
1) Представим десятичные дроби в виде обыкновенных:
$$0{,}42=\frac{42}{100}=\frac{21}{50}$$
$$0{,}072=\frac{72}{1000}=\frac{9}{125}$$
$$4{,}03=4\frac{3}{100}$$
$$6{,}125=6\frac{125}{1000}=6\frac{1}{8}$$
2) Вычислим значения выражений:
$$18\frac{5}{12}-\frac{7}{12}\cdot 1\frac{19}{21}-\frac{17}{12}\cdot \frac{2}{3}$$
$$=18\frac{5}{12}-\frac{7}{12}\cdot \frac{40}{21}-\frac{17}{12}\cdot \frac{2}{3}$$
$$=18\frac{5}{12}-\frac{10}{9}-\frac{17}{18}$$
$$=17\frac{33}{108}-\frac{17}{108}=17\frac{16}{27}$$
$$\left(6\frac{3}{4}-5\frac{1}{8}:\,1\frac{9}{32}\right)\cdot \frac{5}{11}$$
$$= \left(6\frac{3}{4}-\frac{41}{8}:\frac{41}{32}\right)\cdot \frac{5}{11}$$
$$= \left(6\frac{3}{4}-4\right)\cdot \frac{5}{11}$$
$$=2\frac{3}{4}\cdot \frac{5}{11}=\frac{11}{4}\cdot \frac{5}{11}=\frac{5}{4}=1\frac{1}{4}$$
$$(-1{,}42-(-3{,}22)):(-0{,}4)+(-6)\cdot(-0{,}7)$$
$$=( -1{,}42+3{,}22):(-0{,}4)+4{,}2$$
$$=1{,}8:(-0{,}4)+4{,}2=-4{,}5+4{,}2=-0{,}3$$
$$\left(-\frac{7}{18}+\frac{11}{22}\right):\left(-\frac{19}{48}\right)$$
$$=\left(-\frac{7}{18}+\frac{1}{2}\right):\left(-\frac{19}{48}\right)$$
$$=\left(-\frac{14}{36}+\frac{18}{36}\right):\left(-\frac{19}{48}\right)=\frac{4}{36}:\left(-\frac{19}{48}\right)$$
$$=\frac{1}{9}\cdot \left(-\frac{48}{19}\right)=-\frac{16}{57}$$
$$\left(-3\frac{1}{12}-2\frac{1}{15}\right):\left(-5\frac{3}{20}\right)$$
$$=\left(-\frac{37}{12}-\frac{31}{15}\right):\left(-\frac{103}{20}\right)$$
$$=\left(-\frac{185}{60}-\frac{124}{60}\right):\left(-\frac{103}{20}\right)$$
$$=-\frac{309}{60}:\left(-\frac{103}{20}\right)=\frac{309}{60}\cdot \frac{20}{103}=1$$
Ответ
1) $$\frac{21}{50};\ \frac{9}{125};\ 4\frac{3}{100};\ 6\frac{1}{8}$$
2) $$17\frac{16}{27};\ 1\frac{1}{4};\ -0{,}3;\ -\frac{16}{57};\ 1$$
