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HH
2
3 tháng 10 2021
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{2.2}{1.3}\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)
\(=\frac{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
3 tháng 10 2021
\(\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right)^2}< 0\)
\(\Rightarrow\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right).\left(x-2\right)}< 0\)
=> ( x - 3 ) . ( x - 5 ) và ( x - 2 ) . ( x - 2 ) trái dấu
Mà ( x - 2 )2 = ( x - 2 ) . ( x - 2 ) ≥ 0 ∀ x
\(\Rightarrow\hept{\begin{cases}\left(x−3\right).\left(x+5\right)< 0\\\left(x−2\right).\left(x−2\right)>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5;−5< x< 3\\x>2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5\\2< x< 3\end{cases}}\)
KC
0
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)
=>x-300=0
hay x=300
L
0
20 tháng 4 2019
a,A=\(\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2016.2016}{2015.2017}\right)=\frac{1}{2}.\left(\frac{2.3.4...2016}{1.2....2015}.\frac{2.3.4...2016}{3.4....2017}\right)=\frac{1}{2}.\left(\frac{2016.2}{2017}\right)=\frac{4032}{4034}=\frac{2016}{2017}\)
Hok tốt
20 tháng 4 2019
\(\left|x\right|=\frac{1}{2}\Rightarrow x=\orbr{\begin{cases}\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
TH1:\(x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{3}{2}+5=4\)
TH2:\(x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}+\frac{3}{2}+5=7\)
Vậy
TQ
1
18 tháng 3 2023
\(P=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{2021.2023}\)
Ta sẽ "tách" P làm 2 phần:
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)
\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
Do đó \(P=A+B\)
Ta có \(A=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1011}{2023}\)
Mặt khác, \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+\dfrac{8-6}{6.8}+...+\dfrac{2022-2020}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{505}{2022}\)
Từ đó \(P=A+B=\dfrac{1011}{2023}+\dfrac{505}{2022}=\dfrac{3065857}{4090506}\)
