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\(A=\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+...+\frac{1}{168}\)
\(\frac{1}{3}A=\frac{1}{54}+\frac{1}{108}+...+\frac{1}{504}\)
\(\frac{1}{3}A=\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{21.24}\)
\(=\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{21}-\frac{1}{24}\)
\(=\frac{1}{6}-\frac{1}{24}\)
\(=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}\)
=> \(A=\frac{1}{8}:\frac{1}{3}\)\(=\frac{3}{8}\)
a. =(1+10)×10 :2
=11×10:2
=110:2
=55
b. Số các số hạng là =(90-9):9+1= 10
Tổng = (9+90)×10:2=495
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
a)
Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:
2/9+6/27+8/36+12/54+16/72+18/81=
2/9+2/9+2/9+2/9+2/9+2/9=
2/9*6=
12/9=
4/3
Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3
b)
Ta có:
1-2/5=3/5
1-2/7=5/7
1-2/9=7/9
...
1-2/99=97/99
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=
3/5*5/7*7/9*...*97/99=
(3*5*7*...*97)/(5*7*9*...*99)=
3/99=
1/33
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33
c)
Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:
S=1/2+1/4+1/8+1/16+...+1/1024
S*2=1+1/2+1/4+1/8+...+1/512
S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)
S=1-1/1024
S=1023/1024
Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024
\(A=\frac{1}{1\times6}+\frac{1}{6\times11}+\frac{1}{11\times16}+...+\frac{1}{31\times36}\)
\(=\frac{1}{5}.\left(\frac{5}{1\times6}+\frac{5}{6\times11}+...+\frac{5}{31\times36}\right)=\frac{1}{5}\times\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(=\frac{1}{5}\times\left(1-\frac{1}{36}\right)=\frac{1}{5}\times\frac{35}{36}=\frac{7}{36}\)
\(\frac{1\cdot3\cdot9+2\cdot6\cdot18+3\cdot9\cdot27}{1\cdot5\cdot18+2\cdot10\cdot36+3\cdot15\cdot54}\)
\(=\frac{1\cdot3\cdot9+2\left(1\cdot3\cdot9\right)+3\left(1\cdot3\cdot9\right)}{1\cdot5\cdot18+2\left(1\cdot5\cdot18\right)+3\left(1\cdot5\cdot18\right)}\)
\(=\frac{\left(1\cdot3\cdot9\right)\left(1+2+3\right)}{\left(1\cdot5\cdot18\right)\left(1+2+3\right)}\)
\(=\frac{3}{10}\)
a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
ko hiểu chi hết mà đưa ra cái sồm
ko phải tui ra đề đâu đề thi của trường chuyên vĩnh yên cấp 2 do sở ra đề