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Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
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\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
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a) \(\left(3\frac{1}{2}-2x\right).3\frac{1}{3}=7\frac{1}{3}\)
\(\left(\frac{7}{2}-2x\right).\frac{10}{3}=\frac{22}{3}\)
\(\frac{7}{2}-2x=\frac{11}{5}\)
\(2x=\frac{13}{10}\)
\(x=\frac{13}{20}\)
Vậy ...
b) \(\frac{4}{9}x=\frac{9}{8}-0,125\)
\(\frac{4}{9}x=1\)
\(x=\frac{9}{4}\)
Vậy...
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\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
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\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)
\(x\times4+\frac{40}{81}=\frac{51}{81}\)
\(x\times4=\frac{51}{81}-\frac{40}{81}\)
\(x\times4=\frac{11}{81}\)
\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{11}{324}\)
[ 61 + ( 53 - x ) ] . 17 = 1785
61 + ( 53 - x ) = 1785 : 17
61 + ( 53 - x ) = 105
( 53 - x ) = 105 - 61
53 - x = 44
=> x = 53 - 44
=> x = 9
\(\frac{1}{3^x}.\frac{1}{9}.\frac{1}{81}=\frac{1}{3^{20}}\)
\(\Rightarrow\frac{1}{3^x}=\frac{1}{3^{20}}:\left(\frac{1}{3^2}.\frac{1}{3^4}\right)\)
\(\Rightarrow\frac{1}{3^x}=\frac{1}{3^{14}}\)
\(\Rightarrow x=14\)
\(\frac{1}{3^x}.\frac{1}{3^2}.\frac{1}{3^4}=\frac{1}{3^{14}}.\frac{1}{3^2}.\frac{1}{3^4}\)
Vậy x = 14
Study well######