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TD
0
LK
0
6 tháng 5 2016
\(S=\frac{101}{102}+\frac{1}{1.2.2.3}+\frac{1}{2.3.2.3}+\frac{1}{3.4.2.3}+...+\frac{1}{17.18.2.3}=\frac{101}{102}+\frac{1}{6}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)
Đặt BT trong ngoặc đơn là A
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{18-17}{17.18}=1-\frac{1}{18}=\frac{17}{18}\)
\(S=\frac{101}{120}+\frac{1}{6}.\frac{17}{18}\)
8 tháng 5 2017
1.
A=\(\dfrac{3\left|x\right|+2}{\left|x\right|-5}=\dfrac{3\left|x\right|-15+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)}{\left|x\right|-5}+\dfrac{17}{\left|x-5\right|}=3+\dfrac{17}{\left|x\right|-5}\)
Để A \(\in\)Z thì \(\left|x\right|-5\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)
Ta có :
\(\left|x\right|-5=-17\Rightarrow\left|x\right|=-12\left(KTM\right)\)
\(\left|x\right|-5=-1\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\left|x\right|-5=1\Rightarrow\left|x\right|=6\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(\left|x\right|-5=17\Rightarrow\left|x\right|=32\Rightarrow\left[{}\begin{matrix}x=32\\x=-32\end{matrix}\right.\)
Vậy để A \(\in\)Z thì x \(\in\) {-32;-6;-4;4;6;32}
TQ
1
DD
Đoàn Đức Hà
Giáo viên
11 tháng 5 2021
\(A=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{n\left(n+5\right)}\)
\(A=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{n\left(n+5\right)}\right)\)
\(A=\frac{1}{5}\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{n+5-n}{n\left(n+5\right)}\right)\)
\(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+5}\right)\)
\(A=\frac{1}{5}\left(1-\frac{1}{n+5}\right)\)
\(A=\frac{n+4}{5n+25}\)
DD
Đoàn Đức Hà
Giáo viên
11 tháng 5 2021
\(B=1.2+2.3+3.4+...+n\left(n+1\right)\)
\(3B=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3\)
\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(3B=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)\)
\(3B=n\left(n+1\right)\left(n+2\right)\)
\(B=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
12 tháng 4 2016
Ta có: B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<1 ( Vì 172009+1< 172010+1 )
Nên B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<\(\frac{17^{2009}+1+16}{17^{2010}+1+16}\)
=\(\frac{17^{2009}+17}{17^{2010}+17}\)
=\(\frac{17\left(17^{2008}+1\right)}{17\left(17^{2009}+1\right)}\)
=\(\frac{17^{2008+1}}{17^{2009}+1}\)=A
Vậy A>B
\(2023-\dfrac{1}{2\cdot6}-\dfrac{1}{4\cdot9}-\dfrac{1}{6\cdot12}-...-\dfrac{1}{38\cdot60}\)
\(=2023-\left(\dfrac{1}{2\cdot3\cdot\left(1\cdot2\right)}+\dfrac{1}{2\cdot3\cdot\left(2\cdot3\right)}+...+\dfrac{1}{2\cdot3\cdot\left(19\cdot20\right)}\right)\)
\(=2023-\dfrac{1}{6}\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)
\(=2023-\dfrac{1}{6}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2023-\dfrac{1}{6}\left(1-\dfrac{1}{20}\right)=2023-\dfrac{1}{6}\cdot\dfrac{19}{20}\)
\(=2023-\dfrac{19}{120}=\dfrac{242741}{120}\)
023−2⋅61−4⋅91−6⋅121−...−38⋅601
\(= 2023 - \left(\right. \frac{1}{2 \cdot 3 \cdot \left(\right. 1 \cdot 2 \left.\right)} + \frac{1}{2 \cdot 3 \cdot \left(\right. 2 \cdot 3 \left.\right)} + . . . + \frac{1}{2 \cdot 3 \cdot \left(\right. 19 \cdot 20 \left.\right)} \left.\right)\)
\(= 2023 - \frac{1}{6} \left(\right. \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + . . . + \frac{1}{19 \cdot 20} \left.\right)\)
\(= 2023 - \frac{1}{6} \left(\right. 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{19} - \frac{1}{20} \left.\right)\)
\(= 2023 - \frac{1}{6} \left(\right. 1 - \frac{1}{20} \left.\right) = 2023 - \frac{1}{6} \cdot \frac{19}{20}\)
\(= 2023 - \frac{19}{120} = \frac{242741}{120}\)