1.Tìm chữ số A và B biết:\(\dfrac{ }{25a89b}\)chia hết cho 2 và 9 và chia cho 5 thì dư 3
2.Để đánh số trang 1 cuốn sách có 168 trang.Cần dùng bao nhiêu lượt các chữ số?Bao nhiêu các chữ số 5?
3.Tính
a)A=\(\dfrac{1}{3\cdot5}\) +\(\dfrac{1}{5\cdot7}\) +........+\(\dfrac{1}{2017\cdot2019}\)
b)B=\(\dfrac{1}{6}\) +\(\dfrac{1}{12}\) +\(\dfrac{1}{20}\) +...+\(\dfrac{1}{132}\)
Giúp mk vs nha mk đg cần gấp
Bài 3 :
a) \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...........+\dfrac{1}{2017.2019}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+.........+\dfrac{2}{2017.2019}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2019}\)
\(\Leftrightarrow2A=\dfrac{672}{2019}\)
\(\Leftrightarrow A=\dfrac{336}{2019}\)
b) \(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+.........+\dfrac{1}{132}\)
\(\Leftrightarrow B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+............+\dfrac{1}{11.12}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+......+\dfrac{1}{11}-\dfrac{1}{12}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{5}{12}\)
1.
Để \(\overline{25a89b}⋮2\Rightarrow b\in\left\{0;2;4;6;8\right\}\)
Để \(\overline{25a89b}\) chia 5 dư 3 \(\Rightarrow b\in\left\{3;8\right\}\)
Để thỏa mãn hai điều kiện trên thì \(b=8\)
Để \(\overline{25a898}⋮9\Rightarrow\left(2+5+a+8+9+8\right)⋮9\Leftrightarrow32+a⋮9\Rightarrow a=4\)
Vậy \(a=4;b=8\); số cần tìm là \(254898\)
3) a) \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(\Rightarrow2A=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\right)\)
\(2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2A=\dfrac{1}{3}-\dfrac{1}{2019}=\dfrac{224}{673}\Leftrightarrow A=\dfrac{\dfrac{224}{673}}{2}=\dfrac{224}{673.2}=\dfrac{224}{1346}=\dfrac{112}{673}\)
b) \(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{132}\)
\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{11.12}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{12}\)
\(B=\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{5}{12}\)
3.
\(A=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2017\cdot2019}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2017\cdot2019}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2019}\right)\\ =\dfrac{1}{2}\cdot\dfrac{224}{673}\\ =\dfrac{112}{673}\)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{132}\\ =\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{2}-\dfrac{1}{12}\\ =\dfrac{5}{12}\)
