Упр.45 ГДЗ Муравин 7 класс (Алгебра)

1. Найдите значение выражения: 1) $$8a+120$$ при $$a=-9$$; 2) $$5c-111$$ при $$c=-4$$; 3) $$5{,}71x-0{,}71x$$ при $$x=0{,}09$$; 4) $$1{,}42y+0{,}38y$$ при $$y=-100$$; 5) $$\frac{13}{16}-\left(c-\frac{3}{4}\right)$$ при $$c=\frac{9}{16}$$; 6) $$2b+\frac{5}{21}-\frac{17}{35}$$ при $$b=\frac{11}{35}$$; 7) $$12(2x+3y)$$ при $$x=\frac{2}{3}$$, $$y=\frac{3}{4}$$; 8) $$(14a-63b):7$$ при $$a=\frac{1}{2}$$, $$b=-\frac{1}{18}$$.

1. При $$a=-9$$:

   $$8a+120=8\cdot(-9)+120=-72+120=48.$$

2. При $$c=-4$$:

   $$5c-111=5\cdot(-4)-111=-20-111=-131.$$

3. При $$x=0{,}09$$:

   $$5{,}71x-0{,}71x=5x=5\cdot0{,}09=0{,}45.$$

4. При $$y=-100$$:

   $$1{,}42y+0{,}38y=1{,}8y=1{,}8\cdot(-100)=-180.$$

5. При $$c=\frac{9}{16}$$:

   $$\frac{13}{16}-\left(c-\frac{3}{4}\right) =\frac{13}{16}-\left(\frac{9}{16}-\frac{3}{4}\right) =\frac{13}{16}-\left(\frac{9}{16}-\frac{12}{16}\right) =\frac{13}{16}+\frac{3}{16}=1.$$

6. При $$b=\frac{11}{35}$$:

   $$2b+\frac{5}{21}-\frac{17}{35} =2\cdot\frac{11}{35}+\frac{5}{21}-\frac{17}{35} =\frac{22}{35}-\frac{17}{35}+\frac{5}{21} =\frac{5}{35}+\frac{5}{21} =\frac{1}{7}+\frac{5}{21} =\frac{3}{21}+\frac{5}{21} =\frac{8}{21}.$$

7. При $$x=\frac{2}{3},\ y=\frac{3}{4}$$:

   $$12(2x+3y)=12\left(2\cdot\frac{2}{3}+3\cdot\frac{3}{4}\right) =12\left(\frac{4}{3}+\frac{9}{4}\right) =12\left(\frac{16}{12}+\frac{27}{12}\right) =12\cdot\frac{43}{12}=43.$$

8. При $$a=\frac{1}{2},\ b=-\frac{1}{18}$$:

   $$(14a-63b):7 =\left(14\cdot\frac{1}{2}-63\cdot\left(-\frac{1}{18}\right)\right):7 =\left(7+\frac{63}{18}\right):7 =\left(7+\frac{7}{2}\right):7 =\frac{21}{2}:7 =\frac{21}{2}\cdot\frac{1}{7} =\frac{3}{2}=1{,}5.$$

Ответ

1) $$48$$; 2) $$-131$$; 3) $$0{,}45$$; 4) $$-180$$; 5) $$1$$; 6) $$\frac{8}{21}$$; 7) $$43$$; 8) $$1{,}5$$.