- Tính tổng :D = 10.12 +11.13+12.14+13.15+....+50.52
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Ta có:\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}...\frac{14^2}{13.15}=\frac{8^2.9^2.....14^2}{7.9.8.10.9.11....13.15}\)
\(=\)\(\frac{\left(8.9.10...14\right)\left(8.9.10...14\right)}{\left(7.8.9...13\right).\left(9.10.11...15\right)}\)
\(=\frac{14.8}{7.15}=\frac{2.7.8}{7.15}=\frac{2.8}{15}=\frac{16}{15}\)
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}.\frac{11^2}{10.12}.\frac{12^2}{11.13}.\frac{13^2}{12.14}.\frac{14^2}{13.15}\)
\(\frac{8^2.9^2.10^2.11^2.12^2.13^2.14^2}{7.9.8.10.9.11.10.12.11.13.12.14.13.15}\)
\(\frac{8.9.10.11.12.13.14}{7.9.10.11.12.13.15}=\frac{8.14}{7.15}=\frac{112}{105}=\frac{16}{15}\)
Học tốt@_@
b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6
=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52
=48*50*52
=>B=20800
d: 9D=1*4*9+4*7*9+...+46*49*9
=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43
=1*2*4+46*49*52
=117216
=>D=13024
a:
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
Đặt \(A=\frac{3}{10.12}+\frac{3}{12.14}+.....+\frac{3}{48.50}\)
\(A=\frac{3}{2}.\left(\frac{2}{10.12}+\frac{2}{12.14}+......+\frac{2}{48.50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+....+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\frac{2}{25}\)
\(A=\frac{3}{25}\)
S = \(\dfrac{1}{10.12}\) + \(\dfrac{1}{12.14}\) + .....+ \(\dfrac{1}{998.1000}\)
S = \(\dfrac{1}{2}\).( \(\dfrac{2}{10.12}\) + \(\dfrac{2}{12.14}\)+.....+ \(\dfrac{2}{998.1000}\))
S = \(\dfrac{1}{2}\).( \(\dfrac{1}{10}\)- \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{14}\)+...+\(\dfrac{1}{998}\)- \(\dfrac{1}{1000}\))
S = \(\dfrac{1}{2}\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{1000}\))
S = \(\dfrac{1}{2}\).\(\dfrac{99}{1000}\)
S = \(\dfrac{99}{2000}\)