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Задача Задача №3652651
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Аноним
14 сентября 2026 02:32 · публичный вопрос
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Аноним

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Краткие ответы
В №4 знак после \(0{,}36\) читаю как знак процента.

  1. \(\displaystyle \frac{7\,259\,381}{1\,410\,300}\approx5{,}1474\)
  2. \(\displaystyle \frac{28}{3}=9\frac13\)
  3. \(8{,}432\)
  4. \(\displaystyle \frac{3159}{700}=4\frac{359}{700}\)
  5. \(42\)
  6. \(10\)
  7. \(\displaystyle \frac53\)
  8. \(\displaystyle \frac23\)
  9. \(\displaystyle \frac{11}{5}=2\frac15\)
  10. \(\displaystyle -\frac{30}{17}=-1\frac{13}{17}\)
  11. \(81\)

Решение

1)
\[ (20{,}86:18)+(45:0{,}36):(19{,}39+11{,}95) \]

\[ 20{,}86:18=\frac{1043}{900},\qquad 45:0{,}36=125, \] \[ 19{,}39+11{,}95=31{,}34=\frac{1567}{50}. \]

Тогда \[ \frac{1043}{900}+125:\frac{1567}{50} =\frac{1043}{900}+\frac{6250}{1567} =\frac{7\,259\,381}{1\,410\,300} \approx5{,}1474. \]

2)
\[ \frac7{36}\cdot9+8\cdot\frac{11}{12} +\frac9{10}\cdot\frac5{18} \] \[ =\frac74+\frac{22}{3}+\frac14 =2+\frac{22}{3} =\frac{28}{3}=9\frac13. \]

3)
\[ \left(3\frac4{25}+0{,}24\right)\cdot2{,}13 +\left(5{,}1625-2\frac3{16}\right)\cdot\frac25 \]

\[ 3\frac4{25}+0{,}24=3{,}16+0{,}24=3{,}4, \] \[ 3{,}4\cdot2{,}13=7{,}242. \]

\[ 2\frac3{16}=2{,}1875,\qquad 5{,}1625-2{,}1875=2{,}975, \] \[ 2{,}975\cdot\frac25=1{,}19. \]

\[ 7{,}242+1{,}19=8{,}432. \]

4)
\[ 0{,}36\%:\frac7{25}+\frac5{16}:0{,}125 +2\frac12\cdot0{,}8 \]

\[ 0{,}36\%=0{,}0036=\frac9{2500}, \] \[ \frac9{2500}:\frac7{25}=\frac9{700}, \qquad 0{,}125=\frac18, \] \[ \frac5{16}:\frac18=\frac52, \qquad 2\frac12\cdot0{,}8=2. \]

\[ \frac9{700}+\frac52+2 =\frac9{700}+\frac{3150}{700} =\frac{3159}{700}. \]

5)
\[ \sqrt{63}\cdot\sqrt{28} =3\sqrt7\cdot2\sqrt7 =6\cdot7=42. \]

6)
\[ \sqrt{20}\cdot\sqrt5 =2\sqrt5\cdot\sqrt5=10. \]

7)
\[ \sqrt{50}:\sqrt{18} =\frac{5\sqrt2}{3\sqrt2} =\frac53. \]

8)
\[ \sqrt{12}:\sqrt{27} =\frac{2\sqrt3}{3\sqrt3} =\frac23. \]

9)
\[ \left(\frac12+\frac27\right):\frac5{14} =\left(\frac7{14}+\frac4{14}\right)\cdot\frac{14}{5} \] \[ =\frac{11}{14}\cdot\frac{14}{5} =\frac{11}{5}=2\frac15. \]

10)
\[ 1\frac8{12}:\left(1\frac{15}{18}-2\frac79\right) \]

\[ 1\frac8{12}=\frac53,\qquad 1\frac{15}{18}=\frac{11}{6},\qquad 2\frac79=\frac{25}{9}. \]

\[ \frac{11}{6}-\frac{25}{9} =\frac{33}{18}-\frac{50}{18} =-\frac{17}{18}. \]

\[ \frac53:\left(-\frac{17}{18}\right) =\frac53\cdot\left(-\frac{18}{17}\right) =-\frac{30}{17}. \]

11)
\[ \frac{15\cdot2\left(\frac32-\frac24\right)}{\frac36} +\frac{15}{\left(\frac{10}{6}-\frac23\right)}:\frac57 \]

\[ \frac32-\frac24=1,\qquad \frac{15\cdot2\cdot1}{1/2}=60. \]

\[ \frac{10}{6}-\frac23=\frac53-\frac23=1, \] \[ \frac{15}{1}:\frac57=15\cdot\frac75=21. \]

\[ 60+21=81. \]