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\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4072323}{2018\cdot2018}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2017\cdot2019\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(2018\cdot2018\right)}\)
\(-A=\frac{\left(1\cdot2\cdot...\cdot2017\right)\left(3\cdot4\cdot...\cdot2019\right)}{\left(2\cdot3\cdot...\cdot2018\right)\left(2\cdot3\cdot...\cdot2018\right)}\)
\(-A=\frac{1\cdot2019}{2018\cdot2}\)
\(-A=\frac{2019}{4036}\)
\(A=-\frac{2019}{4036}< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right)\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right)...\left(\frac{1}{2018}-\frac{2018^2}{2018^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-4072323}{2018^2}\)
\(\Rightarrow\frac{-\left(3.8.....4072323\right)}{\left(2.3.4...2018\right).\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(1.3.2.4....2017.2019\right)}{\left(2.3.4...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(\left(1.2.3...2017\right).\left(3.4.5..2019\right)\right)}{\left(2.3...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-2019}{2018.2}< -\frac{2018}{2018.2}=\frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\)
P/s: mk ko copy baì của bn uyên đâu nha
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
\(A=\frac{\left(1-2^2\right)\left(1-3^2\right)\left(1-4^2\right)...\left(1-2018^2\right)}{2^23^24^2...2018^2}\)
\(A=\frac{-1\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot...\cdot\left(-2016\right)\cdot2018}{2018!^2}\)
\(A=\frac{2016!\cdot3\cdot4\cdot5\cdot...\cdot2018}{2018!^2}=\frac{2016!\cdot2018!}{2018!^2\cdot2!}=\frac{2016!}{2!2018!}=\frac{1}{2!\cdot2017\cdot2018}>0>-\frac{1}{2}=B\)
A = (1/2+1)(1/2-1)(1/3+1)(1/3-1)....(1/2018+1)(1/2018-1) đặt các tích phần tử có dấu + là X, tích các phần tử có dấu - là Y => A= X.Y
X = 3/2.4/3.5/4.....2019/2018 = 2019/2
Y= (-1/2)(-2/3)(-3/4)...(-2017/2018) = -1/2018 (tích của 2017 số <0)
A= X.Y = -2019/2018.1/2 < B= -1/2
\(D=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{99.101}{100^2}\)
\(=\frac{1.2...99}{2.3...100}.\frac{3.4....101}{2.3....100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
1 b) Đặt A=\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{66}+\frac{1}{78}\)
=> \(\frac{A}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}+\frac{1}{156}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{11.12}+\frac{1}{12.13}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}=\frac{1}{3}-\frac{1}{13}\)
=> \(A=\frac{2}{3}-\frac{2}{13}\)\(=\frac{20}{39}\)
Ta có: \(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+...+\frac{x}{78}=\frac{220}{39}\)
<=> \(x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{78}\right)=\frac{220}{39}\Leftrightarrow x.\frac{20}{39}=\frac{220}{39}\Leftrightarrow x=11\)
2. So sánh A và B
b) A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{20}\right)\)
A = \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{20}{20}-\frac{1}{20}\right)\)
A = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{18}{19}.\frac{19}{20}\)
A = \(\frac{1.2.3.....19}{2.3.4.....20}\)
A = \(\frac{1}{20}\)
Mà \(\frac{1}{20}\)> \(\frac{1}{21}\)
=> A > B