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6A=2.4.6+4.6.(8-2)+.........+50.52.(54-48)
6A=2.4.6+4.6.8-2.4.6+............+50.52.54-48.50.52
6A=50.52.54
A=50.52.54:6
A=23400
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(A=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\frac{99}{100}=\frac{99}{40}\)
\(\frac{3}{6.8}+\frac{3}{8.10}+.......+\frac{3}{198.200}\)
\(=\frac{3}{2}.\left(\frac{2}{6.8}+\frac{2}{8.10}+........+\frac{2}{198.200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+........+\frac{1}{198}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\frac{97}{600}=\frac{97}{400}\)
\(3.\left(\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{198.200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{198}{200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{200}\right).\frac{1}{2}\)
=.\(3.\frac{97}{600}.\frac{1}{2}\)=97/400
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$
\(A=\frac{2}{3}+\frac{2}{15}+...+\frac{2}{143}\)
\(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{11\cdot13}\)
\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(A=1-\frac{1}{13}=\frac{12}{13}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
1/1 ; 1/2 ; 2/1 ; 1/3 ; 2/2 ; 3/1 ; 1/4 ; 2/2 ; 3/2 ; 4/1 Quy luat: tông cua tư sô va mâu sô cua môi phân sô lân lươt la cac so tu nhiên liên tiêp 1;2;3;4;5;.. tư sô băt đâu tư 1 va tăng dân
a) 1 và $\frac{2}{5}$
$1 = \frac{1}{1} = \frac{{1 \times 5}}{{1 \times 5}} = \frac{5}{5}$
Ta có $\frac{5}{5}$ và $\frac{2}{5}$
b) 2 và $\frac{3}{8}$
$2 = \frac{2}{1} = \frac{{2 \times 8}}{{1 \times 8}} = \frac{{16}}{8}$
Ta có $\frac{{16}}{8}$ và $\frac{3}{8}$
c) $\frac{1}{3}$ và 5
$5 = \frac{5}{1} = \frac{{5 \times 3}}{{1 \times 3}} = \frac{{15}}{3}$
Ta có $\frac{1}{3}$ và $\frac{{15}}{3}$
a: \(1=\dfrac{1}{1}=\dfrac{1\cdot5}{5\cdot5}=\dfrac{5}{5}\)
\(\dfrac{2}{5}=\dfrac{2}{5}\)
b: \(2=\dfrac{2\cdot8}{1\cdot8}=\dfrac{16}{8}\); \(\dfrac{3}{8}=\dfrac{3}{8}\)
c: \(5=\dfrac{5}{1}=\dfrac{5\cdot3}{1\cdot3}=\dfrac{15}{3};\dfrac{1}{3}=\dfrac{1}{3}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}\)
=\(\frac{1}{2}-\frac{1}{8}\)
=\(\frac{3}{8}\)