Đặt \(A=\frac{3}{4}+\frac{3}{4^2}+\frac{3}{4^3}+...+\frac{3}{4^{100}}\)

\(\Rightarrow4A=3+\frac{3}{4}+\frac{3}{4^2}+...+\frac{3}{4^{99}}\)

\(\Rightarrow4A-A=\left(3+\frac{3}{4}+\frac{3}{4^2}+...+\frac{3}{4^{99}}\right)-\left(\frac{3}{4}+\frac{3}{4^2}+\frac{3}{4^3}+...+\frac{3}{4^{100}}\right)\)

hay \(3A=3-\frac{3}{4^{100}}\)

\(\Rightarrow A=\frac{3-\frac{3}{4^{100}}}{3}=1-\frac{1}{4^{100}}\)

a, \(A=3+3^2+3^3+3^4+...+3^{100}\)

\(3A=3^2+3^3+3^4+3^5+...+3^{101}\)

\(2A=3^{101}-3\)

\(A=\frac{3^{101}-3}{2}\)

b, \(A=3+3^2+3^3+3^4+...+3^{100}\)

\(A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

\(A=40\left(3+3^5+...+3^{97}\right)⋮40\)

a,  Tổng trên có 100 số hạng

Mỗi nhóm có 4số vậy có 25 nhóm 

A =(3+3^2+3^3+3^4)+......+(3^97+3^98+3^99+3^100)

A=3.(1+3+9+27)+........+3^97.(1+3+9+27)

A=3.40+.....+3^97.40

A=40.(3+.....+3^97)

b,    Vì 40chia hết cho 40 nên 40.(3+....+3^97) chia hết cho 40

\(\Leftrightarrow\frac{9}{35}:\frac{2x}{3}=\frac{-11}{8}\)

\(\Leftrightarrow\frac{27}{70x}=\frac{-11}{8}\)

\(\Leftrightarrow-770x=216\)

\(\Leftrightarrow x=-\frac{108}{385}\)

Ta có \(\frac{a+b}{a-b}=-4\)

\(\Rightarrow a+b=-4\left(a-b\right)\)

\(\Rightarrow a+b=-4a+4b\)

\(\Rightarrow5a=3b\)

\(\Rightarrow\frac{b}{a}=\frac{5}{3}\)

Trả lời:

(a+b)/((a-b)=-4

-> a+b = -4a +4b

-> 5a= 3b

->b/a= 5/3

giúp mình đi

giúp đi mà

Vì x,y thuộc Z => 2x+1,y-3 thuộc Z

                       => 2x+1,y-3 thuộc Ư(10)={+1;+2;+5;+10}

Ta có bảng :

Vậy................................................................