Tính:
a A=8/9*15/16*24/25*...(100 thừa số)
b B=3/1+3/1+2+....+3/1+2+3+...+100
c C=(1-1/21)*(1-1/28)*....*(1-1/2005003)
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b, 3/5 + 4/7 + 2/8 + 10/25 + 9/21 + 28/16
= 3/5 + 4/7 + 2/8 + 2/5 + 3/7 + 14/8
= (3/5 + 2/5) + ( 4/7 + 3/7) + ( 2/8 + 14/8)
= 1 + 1 + 7/4
= 2 + 7/4 = 15/4
c , 8/7 + 7/6 + 5/8 + 10/12 + 24/28 + 6/16
= c , 8/7 + 7/6 + 5/8 + 5/6 + 6/7 + 1/2
= (8/7 + 6/7) + (7/6 + 5/6) + 5/8 + 1/2
= 14/7 + 12/6 + 5/8 + 1/2
= 2 + 2 + 5/8 + 1/2
= 4 + 9/8 = 41/8
a, 21/11-9/19-3/13-12/38-24/22 =21/11-(9/19-3/13)-12/38-24/22 =207/143-12/38-24/22 =3075/2717-24/22 =111/2717 b,2/3+3/4+4/5+12/9+28/16+48/15 =17/12+4/5+12/9+28/16+48/15 =133/60+12/9+28/16+48/15 =71/20+28/16+48/15 =53/10+48/15 =17/2 c, 4/5-1/2-1/4-1/8-1/16-1/36 =3/10-1/4-1/8-1/36 =1/20-1/8-1/16-1/36 =-3/40-1/16-1/36 =-11/60-1/36=-119/720
A = ((20 + 1) . 20 : 2) . 2 = 420
B = (25 + 20) . 6 : 2 = 135
C = ( 33 + 26) . 8 : 2 = 236
D = (1 + 100) .100 : 2 = 5050
bài 1
a,\((\)\(\dfrac{-4}{21}\)\()\)x =\(\dfrac{28}{3}\)\(\times\)\(\dfrac{3}{28}\)
\(\Leftrightarrow\)\(\dfrac{-4}{21}\) x =1
\(\Rightarrow\)x = \(\dfrac{-21}{4}\)
b, \(\dfrac{17}{33}\)x = \(\dfrac{1}{56}\)\(\times\)56
\(\Leftrightarrow\)\(\dfrac{17}{33}\)x = 1
\(\Rightarrow\)x = \(\dfrac{33}{17}\)
bài 2 :
a, A=\(\dfrac{25}{32}\)
số nghịch đảo của A là \(\dfrac{32}{25}\)
B=\(\dfrac{3}{7}\)
số nghịch đảo của B là \(\dfrac{7}{3}\)
b, gọi tổng hai số nghịch đảo 2 số đó là Q
Q= \(\dfrac{32}{25}\) +\(\dfrac{7}{3}\)=\(\dfrac{271}{75}\)
\(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{20.21}\)
\(M=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(M=3\left(1-\dfrac{1}{21}\right)\)
\(M=3.\dfrac{20}{21}=\dfrac{20}{7}\)
\(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{100}{99}\)
\(N=\dfrac{4.9.16.25...100}{3.8.15.24...99}\)
\(N=\dfrac{2.2.3.3.4.4.5.5...10.10}{1.3.2.4.3.5.4.6...9.11}\)
\(N=\dfrac{2.3.4.5...10}{1.2.3...9}.\dfrac{2.3.4.5...10}{3.4.5...11}\)
\(N=10.\dfrac{2}{11}=\dfrac{20}{11}\)
a) \(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+......+\dfrac{3}{20.21}\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{21}\right)\)
= \(3.\dfrac{20}{21}\)
= \(\dfrac{20}{7}\)
b) \(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.......\dfrac{100}{99}\)
= \(\dfrac{4.9.16.25.....100}{3.8.15.24.....99}\)
= \(\dfrac{2.2.3.3.4.4.5.5.......10.10}{1.3.2.4.3.5.4.6......9.11}\)
= \(\dfrac{\left(2.3.4.5.....10\right).\left(2.3.4.5.....10\right)}{\left(1.2.3.4......9\right).\left(3.4.5.....11\right)}\)
= \(\dfrac{10.2}{1.11}\)
= \(\dfrac{20}{11}\)
a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)