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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
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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
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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
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\(9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
8A= 9-\(\frac{1}{3^{100}}\)
vậy => n=100