Ta có : 

\(S=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)

\(\Leftrightarrow\)\(3S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Leftrightarrow\)\(3S-S=\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\right)\)

\(\Leftrightarrow\)\(2S=\frac{1}{3}-\frac{1}{3^9}\)

\(\Leftrightarrow\)\(2S=\frac{3^8-1}{3^9}\)

\(\Leftrightarrow\)\(S=\frac{3^8-1}{2.3^9}\)

Ở đây mk chỉ ghi \(...\) cho nhanh nếu bạn làm vào vở thì ghi đầy đủ ra nhé 

bạn còn on ko

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^{^2}}+...+\frac{1}{3^{98}}\)

\(\Leftrightarrow3A-A=1-\frac{1}{3^{99}}\)

\(\Leftrightarrow2A=1-\frac{1}{3^{99}}\)

\(\Leftrightarrow A=\left(1-\frac{1}{3^{99}}\right)\div2\)

1:

a)10/20-15/20+16/20=-5/20+16/20

=11/20.

b)2/3+8/3:8/5=2/3+5/3

=7/3.

2:a)Ta có:

2x=1/4=3/4

2x=4/4=1

x=1:2

x=0,5

b)x:(2/12-1/12)=-3/8.

x:1/12=-3/8.

x=-3/8x1/12.

x=-1/32.

a/ \(8^5=\left(2^3\right)^5=2^{15}\)và \(32^3=\left(2^5\right)^3=2^{15}\Rightarrow8^5=32^3\)

b/ \(27^4=\left(3^3\right)^4=3^{12}\) và \(9^6=\left(3^2\right)^6=3^{12}\Rightarrow27^4=9^6\)

c/ \(23^{17}-23^{16}=23^{16}\left(23-1\right)=22.23^{16}\)

\(23^{16}-23^{15}=23^{15}\left(23-1\right)=22.23^{15}\)

\(\Rightarrow22.23^{16}>22.23^{15}\Rightarrow23^{17}-23^{16}>23^{16}-23^{15}\)

d/ \(\frac{3^{2015}+1}{3^{2016}}=\frac{1}{3}+\frac{1}{3^{2016}}\) và \(\frac{3^{2016}+1}{3^{2017}+1}=\frac{3^{2017}+3}{3\left(3^{2017}+1\right)}=\frac{3^{2017}+1+2}{3\left(3^{2017}+1\right)}=\frac{1}{3}+\frac{2}{3}.\frac{1}{3^{2017}+1}\)

\(\frac{1}{3^{2016}}>\frac{1}{3^{2017}}>\frac{1}{3^{2017}+1}>\frac{2}{3}.\frac{1}{3^{2017}+1}\)

\(\Rightarrow\frac{3^{2015}+1}{3^{2016}}>\frac{3^{2016}+1}{3^{2017}+1}\)

Câu cuối phân tích tương tự

Bài 1

a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)

\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)

\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)

\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)

\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)

\(=-\dfrac{24}{33}\)

\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)

Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)

\(\Rightarrow2D=1-\frac{1}{3^8}\)

\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)

a ) Co :

 1/1.2 - 1/2.3 = 2/1.2.3 

 1/2.3 - 1/3.4 = 2/2.3.4

 ...

 1/37.38 - 1/38.39 = 2/37.38.39

=> 2M = 2/1.2.3 + 2/2.3.4 + ... + 2/37.38.39

=> 2M = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/37.38 - 1/38.39

=> 2M = 1/2 - 1/1482

=> 2M = 370/741

=> M = 185/741

B ) A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8

3A = 1 + 1/3 + 1/3^2 + ... + 1/3^7

3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^7 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8 )

2A = 1 - 1/3^8

A = ( 1 - 1/3^8 ) / 2