chào châu chấu là 100 dễ mà

là 100 lấy máy tính mà tính

Ta có :\(9A=9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}\)

\(8A=9A-A=9-\dfrac{1}{3^{100}}\)

Mà \(8A=9-\dfrac{1}{3^n}\Rightarrow n=100\)

Vậy n=100

Minh thiếu

\(8A=9A-A=\left(9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}\right)\\ -\left(1+\dfrac{1}{3^2}+\dfrac{1}{3^{\text{4}}}+...+\dfrac{1}{3^{100}}\right)\)

\(8A=9+1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}-1-\dfrac{1}{3^2}-\dfrac{1}{3^4}-...-\dfrac{1}{3^{100}}\)

\(8A=9+\left(1-1\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{3^{98}}-\dfrac{1}{3^{98}}\right)+\dfrac{1}{3^{100}}\)

\(8A=9-\dfrac{1}{3^{100}}\) Mà theo đề bài \(8A=9-\dfrac{1}{3^n}\)

\(\Rightarrow n=100\)

Vậy n=100

n=100

thank you very  much

A=1+1/3²+1/3⁴+.....+1/3¹⁰⁰

=>3².A=9+1/3+/3²+...+1/3⁹⁸

=>9A-A=(9+1/3+1/3²+....+1/3⁹⁸)-(1+1/3²+1/3⁴+.+1/3¹⁰⁰)

=>8A=9-1/3^100=9-1/3^n

=>1/3^100=1/3^n

=>3^100=3^n

=>n=100

Vay n=100

A=.............

=>\(9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

=>\(9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

=>\(8A=9-\frac{1}{3^{100}}\)

=>n=100

Ta có: x-y=8. (1)

          y-z=10. (2)

Từ (1) và (2) =>x-y+y-z=8+10

=>x-z=18

mà x+z=12

=>x=(18+12):2=15

=>z=-3

thay vào (1),ta có:15-y=8

=>y=7

khi đó x+y+z=15+(-3)+7=19

nhớ tk nha