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\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}\)
\(S=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\frac{32}{99}\)
\(S=\frac{16}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(\frac{-1}{2}\right)^2=\frac{-1}{8}\)
Mà \(\left(5x-1\right)^3=\frac{-1}{8}\)
\(\Leftrightarrow5x-1=\frac{-1}{2}\)
\(\Leftrightarrow5x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{10}\)
Vậy ....
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=>3B=3^2+3^3+3^4+...+3^2006
=>3B-B=3^2006-3
=>2B=3^2006-3
=>2B+3=3^2006
**** cho mik nha!
B = 3 + 32 + ... + 32005
3B = 32 + 33 + ... + 32006
3B - B = 32006 - 3
2B = 32006 - 3
=> 32006 - 3 + 3 = 32006
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.
\(\text{(x−2)(2y+1)=17}\)
\(\text{⇒(x−2)(2y+1)=1.17=17.1=(−1).(−17)=(−17).(−1)}\)
Ta có bảng sau:
x−2 | 1 | 17 | −1 | −17 |
2y+1 | 17 | 1 | −17 | −1 |
x | 3 | 19 | 1 | −15 |
y | 8 | 0 | −9 | −1 |
Vậy \(\hept{\begin{cases}x=3\\y=8\end{cases}\hept{\begin{cases}x=19\\y=0\end{cases}\hept{\begin{cases}x=1\\y=-9\end{cases}\hept{\begin{cases}x=-15\\y=-1\end{cases}}}}}\)
học tốt
Đặt A = 1 -3 + 5 - 7 + ..... + 2001 - 2003 + 2005
= (1 - 3) + (5 - 7) + ... + (2001 - 2003) + 2005
= -2 x 501 + 2005
= -1002 + 2005 = 1003
gọi A=1-3+5-7+..+2001-2003+2005
A=(1-3)+(5-7)