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\(S=\frac{101}{102}+\frac{1}{1.2.2.3}+\frac{1}{2.3.2.3}+\frac{1}{3.4.2.3}+...+\frac{1}{17.18.2.3}=\frac{101}{102}+\frac{1}{6}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)
Đặt BT trong ngoặc đơn là A
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{18-17}{17.18}=1-\frac{1}{18}=\frac{17}{18}\)
\(S=\frac{101}{120}+\frac{1}{6}.\frac{17}{18}\)
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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}
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\(S=\frac{101}{120}+\frac{1}{2.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{19-18}{18.19}+\frac{20-19}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{20}\right)=\frac{101}{120}+\frac{19}{120}=\frac{120}{120}=1\)
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