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Ta có :A = \(\frac{5}{3.4}+\frac{5}{4.6}+\frac{5}{5.8}+...+\frac{5}{40.78}=\frac{5}{2.2.3}+\frac{5}{2.3.4}+\frac{5}{2.4.5}+...+\frac{5}{2.39.40}\)
\(=\frac{5}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\right)=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{40}\right)=\frac{5}{2}.\frac{19}{40}=\frac{19}{16}\)
Bài 1
A= 3.4 + 4.5+ 5.6+ .......+ 58.59 + 69.60
3A = 3.4.3 + 4.5.3+ 5.6.3+ .......+ 58.59.3 +59.60.3
= 3.4.(5-2) + 4.5.(6-3)+ 5.6.(7-4)+ .......+ 58.59.(60-57) +59.60.(61-58)
= 3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+..........+ 58.59.60-57.58.59+ 59.60.61-58.59.60
=2.3.4+ 59.60.61= 215964
A= 215964: 3= 71988
Bài 2:
A = 2.4 +4.6+ 6.8+.........+ 96.98+98.100
6A= 2.4.6+4.6.6+6.8.6+.........+96.98.6+98.100.6
= 2.4.6+ 4.6.(8-2) +6.8.(10-4)+.........+96.98.( 100-94) + 98 .100.( 102 - 96)
= 2.4.6+4.6.8-2.4.6 + 6.8.10 -4.6.8+..........+ 96.98.100-94.96.98+ 98.100.102-96.98.100
= 98 .100 .102= 999600
A= 999600:6= 166600
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
\(B=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(B=\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^2\cdot10\cdot\left(3^2\right)^4}\)
\(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^9\cdot10\cdot3^8}\)
\(B=\dfrac{2^{11}\cdot2^2\cdot3^6+2^{11}\cdot3^6\cdot3^3}{2^{11}\cdot2^3\cdot3^6\cdot3+\dfrac{2^{11}}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3\right)+2^{11}\cdot\dfrac{1}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2\right)}\)
\(B=\dfrac{2^2+3^3}{2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2}\)
\(B=\dfrac{4+27}{8\cdot3+\dfrac{1}{4}\cdot10\cdot9}\)
\(B=\dfrac{31}{24+\dfrac{1}{4}\cdot90}\)
\(B=\dfrac{31}{24+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{48}{2}+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{93}{2}}\)
\(B=31\div\dfrac{93}{2}\)
\(B=31\times\dfrac{2}{93}\)
\(B=\dfrac{2}{3}\)
85=32768
3x47=49152
=>85<3x47
1255=30517578125
257=6103515625
=>1255>257
Cách ngắn hơn nè:
125^5 và 25^7
Ta có: 125^5= (5^3)^5= 5^15
25^7= (5^2)^7= 5^14
Vì 5^15>5^14 nên 125^5>25^7
a) 1/1x5 + ... + 1/21x25
= 4 x (1-1/5 + 1/5 - 1/9 + ... + 1/21 - 1/25)
= 1/4 x (1 - 1/25)
= 1/4 x 24/25
= 6/25
`-1,5 \times (7/3-5/3 \times 4)`
`= -1,5 \times (7/3-20/3)`
`= -1,5 \times (-13/3)`
`= 13/2`