Tìm tích:A=\(\left(\dfrac{3}{429}-\dfrac{1}{1.3}\right).\left(\dfrac{3}{429}-\dfrac{1}{3.5}\right)........\left(\dfrac{3}{429}-\dfrac{1}{119121}\right)\)
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24 tháng 4 2015
(3/429 - 1/1.3)(3/429 - 1/3.5) ... (3/429 - 1/121.123)
= (1/143 - 1/1.3)(1/143 - 1/3.5) ... (1/143 - 1/11.13) ... (1/143 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... (1/11.13 -1/11.13) ... (1/11.13 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... 0 ... (1/11.13 - 1/121.123)
= 0
24 tháng 4 2015
=(1/143-1/1.3)...(1/143-1/121.123)
vì trong tích có thừa số (1/143-1/11.13)=0
nên cả tích =0
LÀM ƠN LIKE CHO MÌNH ĐI
24 tháng 4 2021
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
TA
19 tháng 1 2018
Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)
= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)
= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)
=> A=\(\dfrac{7}{15}\)
Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)
=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)
=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)
=> x\(\left(-\dfrac{2}{15}\right)=0\)
=> x=0
19 tháng 1 2018
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)
<=> \(\dfrac{-2}{15}x=0\)
<=> \(x=0\)
Vậy: \(s=\left\{0\right\}.\)
23 tháng 11 2021
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
16 tháng 6 2023
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)