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\(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
= \(2\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\right)\)
= \(2\left(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}\right)\)
= \(\dfrac{2}{3}+2\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+2\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}\right):2+\dfrac{1}{2}\left(\dfrac{4}{8.12}+\dfrac{4}{12.16}\right)\)
= \(\dfrac{2}{3}+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}\)
= \(\dfrac{127}{96}\)
Đặt BT trên là A
\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)
\(A=2A-A=\frac{4}{3}-\frac{2}{192}=\frac{127}{192}\)
`a)-2/3-2/6+2/12+2/24+2/48+2/96+2/192`
`=-4/6-2/6+1/6+1/12+1/24+1/48+1/96`
`=-5/6+1/12+1/24+1/48+1/96`
`=-10/12+1/12+1/24+1/48+1/96`
`=-9/12+1/24+1/48+1/96`
`=-18/24+1/24+1/48+1/96`
`=-17/24+1/48+1/96`
`=-34/48+1/48+1/96`
`=-33/48+1/96`
`=-66/96+1/96=-65/96`
Đặt \(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{2}A-\frac{1}{4}A=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}A=\frac{127}{384}\)
\(A=\frac{127}{384}.4=\frac{127}{96}\)
Đặt \(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\)
\(2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\)
\(2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\right)\)
\(A=\dfrac{4}{3}-\dfrac{2}{192}\)
\(A=\dfrac{127}{96}\)
Vậy \(A=\dfrac{127}{96}\)
Mình hơn nhầm ở chỗ 98 viết lại thành 96. thông cảm nha mình làm lại cho:
Đặt \(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
\(2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}\)
\(2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{96}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\right)\)
\(A=\dfrac{4}{3}-\dfrac{2}{192}\)
\(A=\dfrac{127}{96}\)
Vậy \(A=\dfrac{127}{96}\)
Đặt tổng trên = A
\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)
\(A.2=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)
\(A.2-A=\left(\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\right)-\left(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\right)\)
\(A=\frac{4}{3}-\frac{2}{192}\)
\(A=\frac{127}{96}\)
\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)
\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)
\(2A-A=\frac{4}{3}-\frac{2}{192}\)
\(A=\frac{4}{3}-\frac{2}{192}=\frac{127}{96}\)
\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}.\)
\(A=2A-A=\frac{4}{3}-\frac{2}{96}=\frac{63}{48}\)
+ 2/6 + 2/3 + 2/12 2/24 + 2/48 + 2/96 + 2/192
= 2/3 x (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
Xét phần in Bracket: (1 +1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
1/2 + 1/4 = 1 1/4 -
1/2 + 1/4 + 1/8 = 1 - 1/8
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 1 - 1/64 = 63/64
Suy ra 1 + 63/64 = 127/64
2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96 + 2/192
= 2/3 x 127/64 = 127/96
127/96 h minh len 10 nha