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( \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\)) x \(y\) = \(\frac{2}{3}\)
( \(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\)) x \(y\) = \(\frac{4}{3}\) (nhân 2 vế lên với 2)
(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)- \(\frac{1}{11}\)) x \(y\)= \(\frac{4}{3}\)
( 1 - \(\frac{1}{11}\)) x \(y\)=\(\frac{4}{3}\)
\(\frac{10}{11}\) x \(y\) =\(\frac{4}{3}\)
\(y\) = \(\frac{4}{3}\): \(\frac{10}{11}\)
\(y\) = \(\frac{4}{3}\)x \(\frac{11}{10}\)
\(y\) =\(\frac{22}{15}\)
2/3.5 + 2/5.7 + 2/7.9 + ... + 2/41.43
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/41 - 1/43
= 1/3 - 1/43
= 40/129
ỦNG HỘ NHA
a,[(4x+28):3+55]:5=35
(4x+28):3+55=175
(4x+28):3=120
4x+28=360
4x=332
x=83
a) [( 4x + 28 ) : 3 + 55] : 5 = 35
( 4x + 28 ) : 3 + 55 = 175
( 4x + 28 ) : 3 = 120
4x + 28 = 360
4x = 332
x = 83
b) \(\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\frac{2}{7}\cdot x=\frac{9}{7}\)
\(x=\frac{9}{2}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}\cdot\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{84}{305}\)
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{97\times99}\)
\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{99-3}{297}\)
\(=\frac{96}{297}=\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{37.39}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{37}-\frac{1}{39}\)
\(=\frac{1}{3}-\frac{1}{39}\)
\(=\frac{13}{39}-\frac{1}{39}\)
\(=\frac{12}{39}=\frac{4}{13}\)
ta có A=1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39
=1/3-1/39
=12/39
ta có : 2b= 2/5.7+2/7.9+.....+2/99.101
2b=1/5-1/7+1/7-1/9+...+1/99-1/101
2b= (1/5-1/101)+(1/7-1/7)+(1/9-1/9)+.....(1/99-1/99)
2b= 1/5-1/101
2b= 96/505
2b= 48/505
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{96}{505}\)
\(=\frac{48}{505}\)
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