Алгебра · 7 класс Задача №3749512
А
А
Реши задание на изображении
- \(48\left(\frac{23}{24}-\frac56\right)=48\cdot\frac18=\mathbf6\)
- \(\frac7{10}:\left(\frac38-\frac15\right)=\frac7{10}:\frac7{40}=\mathbf4\)
- \(\frac{13}{9}\cdot\frac87+\frac{22}{63}=\frac{104+22}{63}=\mathbf2\)
- \(\left(1\frac35+2\frac2{15}\right)\cdot\frac9{56}=\frac{56}{15}\cdot\frac9{56}=\mathbf{\frac35}\)
- \(\frac{17}{9}:\left(\frac29+\frac14\right)=\frac{17}{9}:\frac{17}{36}=\mathbf4\)
- \(1\frac12\cdot\frac5{12}:1\frac14=\frac58:\frac54=\mathbf{\frac12}\)
- \(2\frac13\cdot\frac9{16}:3\frac12=\frac{21}{16}:\frac72=\mathbf{\frac38}\)
- \(1\frac23\cdot\frac{14}{25}:1\frac16=\frac{14}{15}:\frac76=\mathbf{\frac45}\)
- \(\frac7{10}:\left(\frac38-\frac15\right)=\mathbf4\)
- \(\frac{19}{4}:\left(\frac14+\frac37\right)=\frac{19}{4}:\frac{19}{28}=\mathbf7\)
- \(1\frac13:\left(\frac56-3\frac12\right)=\frac43:\left(-\frac83\right)=\mathbf{-\frac12}\)
- \(\frac59:\left(\frac7{13}-\frac19\right)=\frac59:\frac{50}{117}=\mathbf{\frac{13}{10}}\)
- \(\left(\frac{13}{21}+\frac3{14}\right):\frac5{27}=\frac56:\frac5{27}=\mathbf{\frac92}\)
- \(\frac73:\left(\frac59-\frac25\right)=\frac73:\frac7{45}=\mathbf{15}\)
