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a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
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\(A=\dfrac{1}{1\cdot6}-\dfrac{1}{6\cdot11}-\dfrac{1}{11\cdot16}-\dfrac{1}{16\cdot21}-...-\dfrac{1}{46\cdot51}\)
\(=\dfrac{1}{6}-\left(\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+\dfrac{1}{16\cdot21}+...+\dfrac{1}{46\cdot51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{5}{6\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{46\cdot51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{46}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\cdot\dfrac{5}{34}\)
\(=\dfrac{1}{6}-\dfrac{1}{34}\)
\(=\dfrac{7}{51}\)
Vậy \(A=\dfrac{7}{51}\)
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Ta có : \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{(5n+1)(5n+6)}\)
\(=\frac{1}{5}\cdot\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{(5n+1)(5n+6)}\right]\)
\(=\frac{1}{5}\cdot\left[1-\frac{1}{5n+6}\right]=\frac{1}{5}\cdot\frac{5n+6-1}{5n+6}=\frac{1}{5}\cdot\frac{5(n+1)}{5n+6}=\frac{n+1}{5n+6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}-\dfrac{1}{101}\)
\(=\dfrac{101-1}{101}\)
\(=\dfrac{100}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101
= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101
=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)
=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)
=1/5 . (1/1-1/101)
=1/5 . 100/101
= 20/101
5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)
=\(1- {1 \over 101}={100 \over 101}\)
suy ra A =\({20 \over 101}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
bạn ơi hình như đề sai ở chỗ cuối cùng kia kìa chỗ đó có phải : x . x ( 1 + 5 )
Đúng ko bạn ?????
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2}{1\cdot6}+\dfrac{2}{11\cdot16}+...+\dfrac{2}{x\left(x+5\right)}=\dfrac{41}{103}\\ =>\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}\right)=\dfrac{41}{103}\\ =>1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{41}{103}:\dfrac{2}{5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=1-\dfrac{1}{206}\\ =>\dfrac{1}{x+5}=\dfrac{1}{206}\\ =>x+5=206\\ =>x=206-5=201\)