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\(=>\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)-y.\frac{2}{5}=\frac{2}{21}\)
\(=>\left(\frac{1}{5}-\frac{1}{21}\right)-y=\frac{2}{21}:\frac{2}{5}\)
\(=>\frac{16}{105}-y=\frac{5}{21}\)
\(=>y=\frac{16}{105}-\frac{25}{105}\)
\(=>y=\frac{-9}{105}=\frac{-3}{35}\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
Ta có:
A = 1 + 3 + 5 + 7 +... + 101
A = \(\frac{102.51}{2}=2601\)
M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68
M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68
M = (- 2 + - 2 + -2 + ... + - 2 ) + 68
M = 25/2 . ( - 2 ) + 68
M = -25 + 68
M = 43
H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )
H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }
H = ( 1 + 2 + 3 +...+ 99 ) x 0
H = 0
G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302
G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )
G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603
G = -4 + - 4 + -4 + 27 - ... + 603
G = 75 x ( -4 ) + 603
G = -300 + 603
G = 303
2.
a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052
= > \(\frac{100.101}{2}\)+ 2 x X = 5052
= > 5050 + 2 x X = 5052
= > 2X = 2
= > X = 1
1. \(\frac{9}{13}-\frac{11}{26}+\frac{1}{2}=\frac{10}{13}\)
2. \(\frac{1}{5}+\frac{2}{7}+\frac{7}{35}=\frac{24}{35}\)
3. \(\frac{4}{9}+\frac{1}{3}:\frac{3}{8}=\frac{4}{3}\)
4. \(\frac{21}{25}-\frac{1}{5}.\frac{2}{3}=\frac{53}{75}\)