b= 7/3x4+7/4x6+7/5x8+7/6x10+..........+7/60x118
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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)\(\frac{4}{5}\times\frac{8}{3}\times\frac{x}{7}=\frac{96}{105}\)
\(\frac{x}{7}=\frac{96}{105}:\frac{4}{5}:\frac{8}{3}=\frac{3}{7}\)
\(x=3\)
b) \(\frac{7}{4}\times\frac{3}{x}\times\frac{11}{5}=\frac{231}{200}\)
\(\frac{3}{x}=\frac{231}{200}:\frac{7}{4}:\frac{11}{5}=\frac{3}{10}\)
\(x=10\)
\(\frac{3}{5}x\frac{8}{27}x\frac{5}{3}=\frac{3}{3}x\frac{5}{5}x\frac{8}{27}=\frac{8}{27}\)
\(\frac{7}{19}x\frac{1}{3}+\frac{7}{19}x\frac{2}{3}=\frac{7}{19}x\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}\)
\(\dfrac{7}{2\times3}+\dfrac{7}{3\times4}+\dfrac{7}{4\times5}+\dfrac{7}{5\times6}\)
= 7 x ( \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) )
= 7 x \(\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
= 7 x \(\dfrac{1}{3}=\dfrac{7}{3}\)
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
: Diện tích toàn phần của một hình lập phương bằng 486 dm2. Vậy thể tích của hình
lập phương đó là: