\(\dfrac{2}{3}\)+\(\dfrac{2}{54}\)+\(\dfrac{2}{108}\)+...+\(\dfrac{2}{648}\)
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17 tháng 11 2023
\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)
\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)
\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)
\(=\dfrac{4}{x\left(x-12\right)}\)
16 tháng 4 2017
Bài 1:
Ta có:
\(A=\dfrac{1}{3}-\dfrac{1}{18}-\dfrac{1}{54}-\dfrac{1}{108}-\dfrac{1}{270}-\dfrac{1}{378}\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+\dfrac{1}{270}+\dfrac{1}{378}\right)\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{18.21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{18}-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{1}{3}-\dfrac{1}{3}.\dfrac{6}{21}\)
\(=\dfrac{1}{3}-\dfrac{2}{21}=\dfrac{5}{21}\)
Vậy \(A=\dfrac{5}{21}\)
Bài 2:
Ta có: \(51x+26y=2000\)
Mà \(\left\{{}\begin{matrix}26y⋮2\\2000⋮2\end{matrix}\right.\) \(\Leftrightarrow51x⋮2\)
\(\left(51;2\right)=1\Rightarrow x⋮2\)
Mặt khác \(x\) là số nguyên tố nên \(x=2\)
Khi đó:
\(51.2+26y=2000\Leftrightarrow y=73\) (thỏa mãn)
Vậy các số nguyên tố \(\left(x,y\right)=\left(2;73\right)\)
22 tháng 8 2022
a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
26 tháng 5 2022
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
1 tháng 10 2017
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
1 tháng 10 2017
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
GL
24 tháng 8 2023
a) \(\dfrac{2}{5}=\dfrac{2\times3}{5\times3}=\dfrac{6}{15}=\dfrac{2}{5}\)
\(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{8}{14}=\dfrac{4}{7}\)
\(\dfrac{13}{54}=\dfrac{13\times3}{54\times3}=\dfrac{39}{162}=\dfrac{13}{54}\)
b) \(\dfrac{8}{20}=\dfrac{8:4}{20:4}=\dfrac{2}{5}\)
\(\dfrac{10}{16}=\dfrac{10:2}{16:2}=\dfrac{5}{8}\)
\(\dfrac{25}{65}=\dfrac{25:5}{65:5}=\dfrac{5}{13}\)
25 tháng 10 2021
\(a,\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{23}{20}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=-\dfrac{23}{20}\end{matrix}\right.\\ b,\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{2}{9}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{9}\\x+\dfrac{1}{3}=-\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=-\dfrac{5}{9}\end{matrix}\right.\\ c,\Leftrightarrow3^x\cdot6=54\Leftrightarrow3^x=9=3^2\Leftrightarrow x=2\)
\(\dfrac{2}{3}+\dfrac{2}{54}+\dfrac{2}{108}+....+\dfrac{2}{648}\\ =\dfrac{2}{3}+\dfrac{2}{6.9}+\dfrac{2}{9.12}+...+\dfrac{2}{24.27}\\ =\dfrac{2}{3}+\dfrac{2}{3}\left(\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{24.27}\right)\\ =\dfrac{2}{3}+\dfrac{2}{3}.\left(\dfrac{1}{6}-\dfrac{1}{27}\right)\\ =\dfrac{2}{3}+\dfrac{2}{3}.\dfrac{7}{54}\\ =\dfrac{2}{3}+\dfrac{7}{81}\\ =\dfrac{61}{81}\)