Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
⇔ \(2A=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)\)
⇔ 2A = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
⇔ 2A = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
⇔ 2A = \(\dfrac{1}{3}-\dfrac{1}{99}\)
⇔ 2A = \(\dfrac{32}{99}\)
⇔ A = \(\dfrac{32}{99}:2\)
⇔ A = \(\dfrac{16}{99}\)
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{97.99}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{97.99}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{99}\)
\(\Leftrightarrow2A=\dfrac{32}{99}\)
\(\Leftrightarrow A=\dfrac{16}{99}\)
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(A=1-\dfrac{1}{99}\)
\(A=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(B=\dfrac{22}{45}\)
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+.........+\dfrac{2}{97.99}\)
\(\Leftrightarrow A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow A=1-\dfrac{1}{99}=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.......+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+......+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow B=\dfrac{22}{45}\)
b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)
c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?
d)
\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)
\(=1-\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{55\cdot57}\right)\)
\(=1-\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{55}-\dfrac{1}{57}\right)\)
\(=1-\dfrac{1}{2}\cdot\dfrac{18}{57}=1-\dfrac{9}{57}=\dfrac{48}{57}=\dfrac{16}{19}\)
Đặt :
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+........+\dfrac{1}{19.21}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{19.21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.........+\dfrac{1}{19}-\dfrac{1}{21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{21}\)
\(\Leftrightarrow2A=\dfrac{20}{21}\)
\(\Leftrightarrow A=\dfrac{10}{21}\)
Đặt A =
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{19\cdot21}\\ \Rightarrow2A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =1-\dfrac{1}{21}=\dfrac{20}{21}\\ \Rightarrow A=\dfrac{20}{21}:2=\dfrac{10}{21}\)
\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+\left|x+\dfrac{1}{99.101}\right|=100x\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{1.3}\right|\ge0\\\left|x+\dfrac{1}{3.5}\right|\ge0\\\left|x+\dfrac{1}{99.101}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+ \left|x+\dfrac{1}{99.101}\right|\ge0\)\(\Rightarrow100x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+x+\dfrac{1}{5.7}+...+x+\dfrac{1}{99.101}=100x\)\(\Rightarrow50x+\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}=100x\)
\(\Rightarrow50x+\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=100x\)
\(\Rightarrow50x+\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=100x\)
\(\Rightarrow50x+\dfrac{50}{101}=500x\)
\(\Rightarrow50x=\dfrac{50}{101}\)
\(\Rightarrow x=\dfrac{1}{101}\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{8}{17}:\dfrac{1}{2}=\dfrac{16}{17}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{16}{17}=\dfrac{1}{17}\)
\(\Rightarrow x+2=17\rightarrow x=15\)
Vậy x = 15
\(A=\dfrac{1}{3.5} +\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(\Rightarrow2A=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)\)
\(\Rightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(\Rightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Rightarrow2A=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{32}{99}:2=\dfrac{16}{99}\)