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TT
3
1 tháng 4 2016
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
1 tháng 4 2016
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 =\({20 \over 101}\)
TT
2
19 tháng 11 2017
Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)
NN
19 tháng 11 2017
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}-\dfrac{1}{101}\)
\(=\dfrac{101-1}{101}\)
\(=\dfrac{100}{101}\)
7 tháng 4 2017
\(C=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(C=\frac{1}{5}\left(1-\frac{1}{101}\right)\)
\(C=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
7 tháng 4 2017
\(5C=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\)
\(5C=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(5C=1-\frac{1}{101}\)
\(C=\frac{100}{\frac{101}{5}}\)
N
1
9 tháng 8 2019
Ta có : \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{(5n+1)(5n+6)}\)
\(=\frac{1}{5}\cdot\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{(5n+1)(5n+6)}\right]\)
\(=\frac{1}{5}\cdot\left[1-\frac{1}{5n+6}\right]=\frac{1}{5}\cdot\frac{5n+6-1}{5n+6}=\frac{1}{5}\cdot\frac{5(n+1)}{5n+6}=\frac{n+1}{5n+6}\)
BT
1
T
6 tháng 10 2023
\(A=\dfrac{1}{1\cdot6}-\dfrac{1}{6\cdot11}-\dfrac{1}{11\cdot16}-\dfrac{1}{16\cdot21}-...-\dfrac{1}{46\cdot51}\)
\(=\dfrac{1}{6}-\left(\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+\dfrac{1}{16\cdot21}+...+\dfrac{1}{46\cdot51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{5}{6\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{46\cdot51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{46}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{6}-\dfrac{1}{5}\cdot\dfrac{5}{34}\)
\(=\dfrac{1}{6}-\dfrac{1}{34}\)
\(=\dfrac{7}{51}\)
Vậy \(A=\dfrac{7}{51}\)
LN
0
20 tháng 1 2017
a)
=\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5\left(7^3-7^4\right)}{7^3.3^2}\)
=\(\frac{3^4\left(3-1\right)}{^{ }3^4\left(9+3\right)}-\frac{5.7^3-5.7^4}{7^3.3^2}\)
=\(\frac{1}{6}-\frac{7^3.5\left(1-7\right)}{7^3.3^2}=\frac{1}{6}-\frac{30}{9}=-\frac{19}{6}\)
Vậy A=\(-\frac{19}{6}\)
20 tháng 1 2017
câu b lúc nã mk làm sai rui
dây mới đúng
=\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
=\(\frac{1}{5}\left(1-\frac{1}{101}\right)=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
11 tháng 2 2022
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 \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)