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![](https://rs.olm.vn/images/avt/0.png?1311)
Đề: Có ở trên
a) Khi nào nó là 1 phân số?
b) Khi nào nó là một số nguyên?
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{20-19}{19.20}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}$
$=1-\frac{1}{20}=\frac{19}{20}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{2}{7}\cdot\dfrac{-3}{4}\cdot\dfrac{4}{7}=\dfrac{-6}{49}\)
b: \(=\dfrac{3}{5}:\left(\dfrac{-5}{9}\cdot\dfrac{-3}{25}\right)=\dfrac{3}{5}:\dfrac{15}{225}=\dfrac{3}{5}\cdot15=9\)
c: \(=5+\dfrac{6}{7}-2-\dfrac{3}{8}-1-\dfrac{1}{8}=2+\dfrac{6}{7}-\dfrac{1}{2}=\dfrac{33}{14}\)
d: \(=\dfrac{-25}{12}-\dfrac{23}{12}-\dfrac{3}{2}=-4-\dfrac{3}{2}=-\dfrac{11}{2}\)
e: \(=\dfrac{-3}{5}\left(\dfrac{-1}{9}-\dfrac{5}{6}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot\dfrac{14}{9}=\dfrac{-42}{45}=\dfrac{-14}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>2x-4+11 chia hết cho x-2
=>\(x-2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{3;1;13;-9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-4\right)\left(3x+1\right)< 0\)
=> TH1: \(\begin{matrix}2x-4< 0\\3x+1>0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\3x>-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{1}{3}\end{matrix}\right.\) (tm)
TH2: \(\begin{matrix}2x-4>0\\3x+1< 0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>4\\3x< -1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\dfrac{1}{3}\end{matrix}\right.\) (vô lí)
=> \(2>x>-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
=> \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
=> \(C=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{10}-\frac{1}{11}\right)\)
=> \(C=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
m) Có số số hạng là :
[(−201)−(−1)]:(−2)+1=101(số số hạng)
Tổng trên bằng :
[(−201)+(−1)]×101:2=−10201
Đáp số : -10201
n)= 17 + (-20) + 23 + (-26) + ... + 53 + (-56) ( có tất cả 14 số và 7 cặp)
= 17 - 20 + 23 - 26 + ... + 53 - 56
= (17 - 20) + (23 - 26) + ... + (53 - 56)
= -3 + -3 + ... + -3
= -3.7
= -21
Học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 23 + ( -77 ) + ( -23 ) + 77
= [ 23 + ( -23 ) ] + [ ( -77 ) + 77 ]
= 0 + 0 = 0
b, ( -2020 ) + 2021 + 21 + ( -22 )
= ( - 2020 ) + 2021 + 21 + ( -21 ) + ( -1 )
= [ ( -2020 ) + 2021 + ( -1 ) ] + [ 21 + ( -21 ) ]
= 0 + 0 = 0
ở môn nào vậy bạn !
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