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\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{49.50}\)
= \(3.\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{49.50}\right)\)
= \(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{49}-\frac{1}{50}\right)\)
= \(3.\left(1-\frac{1}{50}\right)\)
= \(3.\frac{49}{50}\)
= \(\frac{147}{50}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{49.50}\)
\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=3\left(1-\frac{1}{50}\right)=3.\frac{49}{50}=\frac{199}{50}\)
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Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A = 999900 : 3
A = 333300
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{49.50}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A=1-\(\frac{1}{50}\)
A=\(\frac{49}{50}\)
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A=1.2+2.3+3.4+4.5+...+99.100
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101
=99.100.101=999900
=>A=999900:3=333300
Vậy A=333300
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\(1+1=2\)
\(2.3=6\)
\(8.72:9=576:9=64\)
\(\frac{1234+4321}{4.7+5}=\frac{5555}{28+5}=\frac{5555}{33}=\frac{505.11}{3.11}=\frac{505}{3}\)
\(4,5.5+3,2.\frac{2}{1}=22,5+6,4=28,9\)
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A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)
A=\(1-\frac{1}{4}\)
A=\(\frac{3}{4}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
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y=0 nha bạn toán tiếng anh đúng không
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\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{1}-\frac{1}{20}\)
\(=\frac{19}{20}\)
\(\left(131,4-80,8\right):2,3+21,84\times2\)
\(=50,6:2,3+43,68\)
=22+43,68
=65,68
3354