\(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)

\(=2\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\right)\)

\(=2\times\left(\dfrac{64+32+16+8+4+2+1}{192}\right)\)

\(=\dfrac{127}{96}\)

=2/3+1/3+1/6+1/12+1/24+1/48+1/96

=\(\dfrac{127}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}.\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}=\frac{191}{68}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}\)

\(=\frac{1}{3.1}+\frac{1}{3.2}+\frac{1}{3.2^2}+...+\frac{1}{3.2^6}\)

\(=\frac{1}{3}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(=\frac{1}{3}.A\)với \(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

\(\Rightarrow2A=2.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(\Rightarrow2A=2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\)

\(\Rightarrow2A-A=\left(2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

\(\Rightarrow A=2-\frac{1}{2^6}=2-\frac{1}{64}=\frac{127}{64}\)

\(\Rightarrow\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}=\frac{1}{3}.\frac{127}{64}=\frac{127}{192}\)

`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 tổng trên = A

\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)

\(A.2=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)

\(A.2-A=\left(\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\right)-\left(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\right)\)

\(A=\frac{4}{3}-\frac{2}{192}\)

\(A=\frac{127}{96}\)

\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}+\frac{2}{192}\)

\(2A=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{98}\)

\(2A-A=\frac{4}{3}-\frac{2}{192}\)

\(A=\frac{4}{3}-\frac{2}{192}=\frac{127}{96}\)

Đặt \(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(\frac{1}{2}A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(\frac{1}{4}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)

\(\frac{1}{2}A-\frac{1}{4}A=\frac{1}{3}-\frac{1}{384}\)

\(\frac{1}{4}A=\frac{127}{384}\)

\(A=\frac{127}{384}.4=\frac{127}{96}\)

khong ro moi hoi

\(\frac{1999x1999}{1995x1995_{ }}=\frac{1999^2}{1995^2}=\left(\frac{1999}{1995}\right)^2\)\(>1^2\)\(=1\)

Bài 2 bấm máy tính nhé !

\(\frac{1999\times1999}{1995\times1995}=\frac{1999^2}{1995^2}>1^2\Rightarrow\frac{1999\times1999}{1995\times1995}>1\)

127/96 nhé

nếu sai bạn sửa cho mình nhé

thank you 

mình cảm ơn trước

Bằng 127/96

\(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\\ =1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{2}{192}\\ =\dfrac{5}{4}+\dfrac{7}{96}=\dfrac{127}{96}\)