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a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)
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\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
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\(\dfrac{7}{19}\).\(\dfrac{5}{19}\) + \(\dfrac{7}{19}\).\(\dfrac{8}{13}\) + \(\dfrac{12}{19}\)
= \(\dfrac{7}{19}\).( \(\dfrac{5}{13}+\dfrac{8}{13}\)) + \(\dfrac{12}{19}\)
= \(\dfrac{7}{19}\) . \(\dfrac{13}{13}\) + \(\dfrac{12}{19}\)
= \(\dfrac{7}{19}\) + \(\dfrac{12}{19}\)
= \(\dfrac{19}{19}\)
= 1
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a,Ta có :
1/3 + 1/16 + 1/19 + 1/21 + 1/61 + 1/72 + 1/83 + 1/94 = 0,54
3/4 = 0,6
=>b < 3/4
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Vì \(\frac{19^{30}+5}{19^{31}+5}< 1\); \(\frac{19^{31}+5}{19^{30}+5}>1\)
Nên A < B
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Đặt A = 1 + ( - 2 ) + 3 + ( - 4 ) + ..... + 19 + ( - 20 )
=) Số Số Hạng của A là : ( 20 - 1 ) : 1 + 1 = 20
=) Số Cặp của A là : 20 : 2 = 10
=) A = ( -1 ) + ( -1 ) + ( -1 ) +.......+ ( -1 )
=) A = ( -1 ) . 20 = -20
= [1+(-2)] + [3+(-4)] + [5+ ( -6)] + ... + [19 + ( -20]
= (-1) + (-1) + ... + ( -1) có 20 số -1
= -1. 20
= -20
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Đặt A = 19a + 5b + 1890 x c
+ Với a lẻ => a = 2k+1. Ta có:
A = 192k+1 + 5b + 1890 x c
A = 192k . 19 + (...5) + (...0)
A = (192)k . 19 + (...5)
A = (...1)k . 19 + (...5)
A = (...1) . 19 + (...5)
A = (...9) + (....5) = (....4)
+ Với a chẵn => a = 2k. Ta có:
A = 192k + 5b + 1890 x c
A = (192)k + (....5) + (...0)
A = (...1)k + (....5)
A = (...1) + (...5) = (....6)
Vậy với a lẻ thì 19a + 5b + 1890 x c có tận cùng là 4, với a chẵn thì có tận cùng là 6
Ủng hộ mk nha ^-^
A) \(\frac{50}{31}\)
B) 0
Cảm ơn đã k cho tui