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Minh Tú sét boi
Giới thiệu về bản thân
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\(y\text{×}0,2=1,5\text{×}1,2\)
\(y\text{×}0,2=1,8\)
\(y=1,8:0,2\)
\(y=9\)
\(28\dfrac{17}{100}=\dfrac{28\text{×}100+17}{100}=\dfrac{2817}{100}\)
\(Bài\text{ }giải\)
\(Tổng\text{ }số\text{ }lít\text{ }dầu\text{ }của\text{ }cả\text{ }ba\text{ }thùng\text{ }là:\)
\(30\text{×}3=90\left(lít\text{ }dầu\right)\)
\(Tổng\text{ }số\text{ }lít\text{ }dầu\text{ }của\text{ }thùng\text{ }thứ\text{ }nhất\text{ }và\text{ }thùng\text{ }thứ\text{ }hai\text{ }là:\)
\(40\text{×}2=80\left(lít\text{ }dầu\right)\)
\(Tổng\text{ }số\text{ }lít\text{ }dầu\text{ }của\text{ }thùng\text{ }thứ\text{ }hai\text{ }và\text{ }thùng\text{ }thứ\text{ }ba\text{ }là:\)
\(32\text{×}2=64\left(lít\text{ }dầu\right)\)
\(Số\text{ }lít\text{ }dầu\text{ }của\text{ }thùng\text{ }thứ\text{ }nhất\text{ }là:\) \(90-64=26\left(lít\text{ }dầu\right)\) \(Số\text{ }lít\text{ }dầu\text{ }của\text{ }thùng\text{ }thứ\text{ }hai\text{ }là:\) \(80-26=54\left(lít\text{ }dầu\right)\) \(Số\text{ }lít\text{ }dầu\text{ }của\text{ }thùng\text{ }thứ\text{ }ba\text{ }là:\) \(64-54=10\left(lít\text{ }dầu\right)\) \(Do\text{ }:10< 26< 54\rightarrow Thùng\text{ }thứ\text{ }hai\text{ }chứa\text{ }nhiều\text{ }dầu\text{ }nhất.\)\(-\left(34-25+46\right)+\left[34+\left(-25\right)-16\right]\)
\(=-34+25-46+34-25-16\)
\(=\left(34-34\right)+\left(25-25\right)+\left(-46-16\right)\)
\(=0+0+\left(-62\right)=-62\)
\(Bài\text{ }giải\)
\(Số\text{ }nhãn\text{ }vở\text{ }mà\text{ }Bình\text{ }có\text{ }là:\)
\(12+24=36\left(nhãn\text{ }vở\right)\)
\(Trung\text{ }bình\text{ }cộng\text{ }số\text{ }nhãn\text{ }vở\text{ }của\text{ }Lan\text{ }và\text{ }Bình\text{ }là:\)
\(\left(24+36\right):2=30\left(nhãn\text{ }vở\right)\)
\(Hoa\text{ }có\text{ }số\text{ }nhãn\text{ }vở\text{ }là:\)
\(34+30=64\left(nhãn\text{ }vở\right)\)
\(Đáp\text{ }số:64\text{ }nhãn\text{ }vở.\)
\(a.Chu\text{ }vi\text{ }của\text{ }hình\text{ }chữ\text{ }nhật\text{ }là:\left(5+4\right)\text{×}2=18\left(cm\right)\)
\(b.Chu\text{ }vui\text{ }của\text{ }hình\text{ }vuông\text{ }là:6\text{×}4=24\left(cm\right)\)
\(Diện\text{ }tích\text{ }của\text{ }hình\text{ }vuông\text{ }là:6\text{×}6=36\left(cm^2\right)\)
\(c.Chu\text{ }vi\text{ }của\text{ }hình\text{ }tam\text{ }giác\text{ }là:10+10+5=25\left(cm\right)\)
\(Bài\text{ }giải\)
\(Coi\text{ }giá\text{ }vốn\text{ }của\text{ }chiếc\text{ }quạt\text{ }là:100\%.\)
\(Số\text{ }tiền\text{ }lãi\text{ }bán\text{ }chiếc\text{ }xe\text{ }đạp\text{ }chiếm\text{ }số\text{ }phần\text{ }trăm\text{ }so\text{ }với\text{ }giá\text{ }vốn\text{ }là:\)
\(100\%+25\%=125\%\)
\(Số\text{ }tiền\text{ }vốn\text{ }cửa\text{ }hàng\text{ }mua\text{ }chiếc\text{ }xe\text{ }đạp\text{ }là:\)
\(2500000:125\text{×}100=2000000\left(đồng\right)\)
\(Cửa\text{ }hàng\text{ }phải\text{ }bán\text{ }chiếc\text{ }xe\text{ }đạp\text{ }với\text{ }số\text{ }tiền\text{ }là:\)
\(2000000:100\text{×}\left(100+30\right)=2600000\left(đồng\right)\)
\(Đáp\text{ }số:2600000đồng.\)
\(3419\text{×}102-3419-3419\)
\(=3419\text{×}102-3419\text{×}1-3419\text{×}1\)
\(=3419\text{×}\left(102-1-1\right)\)
\(=3419\text{×}100=341900\)