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A=1/1+5+5^2+5^3+...+5^8+5+5^2+5^3+...+5^9=1/1+5+5^2+5^3+...+5^8+5.
Tương tự B=1/1+3+3^2+...+3^8+3
=>A>B.
k nha.
\(A=\frac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}+\frac{5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{1+5+5^2+....+5^8}=1+\frac{1}{\frac{1+5+5^2+...+5^8}{5^9}}\)
\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}=1+\frac{1}{\frac{1+3+3^2+....+3^8}{3^9}}\)
Nhận xét:
\(\frac{1+5+5^2+...+5^8}{5^9}=\frac{1}{5^9}+\frac{1}{5^8}+\frac{1}{5^7}+...+\frac{1}{5}\); \(\frac{1+3+3^2+...+3^8}{3^9}=\frac{1}{3^9}+\frac{1}{3^8}+\frac{1}{3^7}+....+\frac{1}{3}\)
Vì \(\frac{1}{5^9}<\frac{1}{3^9};\frac{1}{5^8}<\frac{1}{3^8};....;\frac{1}{5}<\frac{1}{3}\)nên \(\frac{1+5+5^2+...+5^8}{5^9}<\frac{1+3+3^2+...+3^8}{3^9}\)
=> \(\frac{1}{\frac{1+5+5^2+...+5^8}{5^9}}>\frac{1}{\frac{1+3+3^2+...+3^8}{3^9}}\)=> A > B
\(A=1+\frac{5^9}{1+5+..+5^8}\)
\(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)
Tương tự:
\(B=1+\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
Vì \(\frac{1}{5}< \frac{1}{3}\) , \(\frac{1}{5^2}< \frac{1}{3^2}\), . . .
nên: \(\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}>\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
=> A > B
Vậy đề bạn cho chứng minh A < B là sai nhé.
Ta có:\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
=>\(A=\frac{\left(1+5+5^2+...+5^8\right)}{\left(1+5+5^2+...+5^8\right)}+\frac{5^9}{1+5+5^2+...+5^8}\)
=>\(A=1+\frac{5^9}{1+5+5^2+...+5^8}\)
Ta có:\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
=>\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}\)
=>\(B=1+\frac{3^9}{1+3+3^2+...+3^8}\)
vì:\(1+3+3^2+...+3^8< 1+5+5^2+...+5^8\)
Nên A<B(đpcm).
Gọi tử là : R
=> \(R=1+5+5^2+5^3+......+5^9\)
\(\Rightarrow5R=5+5^2+5^3+....5^{10}\)
\(\Rightarrow5R-R=5^{10}-1\)
\(\Rightarrow4R=5^{10}-1\)
\(\Rightarrow R=\frac{5^{10}-1}{4}\)
Goij mẫu là M
\(\Rightarrow M=1+5+5^2+5^3+.....+5^8\)
\(\Rightarrow5M=5+5^2+.....+5^9\)
\(\Rightarrow5M-M=5^9-1\)
\(\Rightarrow M=\frac{5^9-1}{4}\)
\(\Rightarrow A=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}=1\)
Tương tự : B
Rồi so sánh thôi dễ mà
Phần B nek :
Gọi tử là : T
\(\Rightarrow T=1+3+3^2+.....+3^9\)
\(\Rightarrow3T=3+3^2+3^3+.....+3^{10}\)
\(\Rightarrow3T-T=3^{10}-1\)
\(\Rightarrow T=\frac{3^{10}-1}{2}\)
Gọi mẫu là : H
\(\Rightarrow H=1+3+3^2+.....+3^8\)
\(\Rightarrow3H=3+3^2+3^3+.....+3^9\)
\(\Rightarrow3H-H=3^9-1\)
\(\Rightarrow H=\frac{3^9-1}{2}\)
\(\Rightarrow B=\frac{T}{H}=\frac{\frac{3^{10}-1}{2}}{\frac{3^9-1}{2}}=\frac{29524}{9841}=3,0001.....\)
Cho a sửa câu a nha :
\(\Rightarrow A=\frac{R}{M}=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}=\frac{2441406}{488281}=5,000002048\)
Vậy \(\Rightarrow A>B\left(đpcm\right)\)
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
Ta có :
\(A=\frac{\left(1+5+5^2+....+5^8\right)+5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{1+5+5^2+...+5^8}=1+1:\frac{1}{\frac{1+5+5^2+....+5^8}{5^9}}\)
\(=1+1:\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+\frac{1}{5^7}+.....+\frac{1}{5}}\)
Tương tự ta cũng có \(B=1+1:\frac{1}{\frac{1+3+3^2+...+3^8}{3^9}}=1+1:\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+\frac{1}{3^7}+....+\frac{1}{3}}\)
Ta thấy : \(\frac{1}{5^9}>\frac{1}{3^9};\frac{1}{5^8}>\frac{1}{3^8};.........;\frac{1}{5}>\frac{1}{3}\)
\(\Rightarrow\frac{1}{5^8}+\frac{1}{5^7}+\frac{1}{5^6}+....+\frac{1}{5}>\frac{1}{3^8}+\frac{1}{3^7}+\frac{1}{3^6}+....+\frac{1}{3}\)
\(\Rightarrow1+1:\frac{1}{\frac{1}{5^8}+\frac{1}{5^7}+\frac{1}{5^6}+....+\frac{1}{5}}>1+1:\frac{1}{\frac{1}{3^8}+\frac{1}{3^7}+\frac{1}{3^6}+.....+\frac{1}{3}}\)
Hay A > B (đpcm)
a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)
b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)
\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)
Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
c, Câu hỏi của truong nguyen kim
\(A=\frac{1+5+5^2+...+5^8+5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{5^8}=6\)
\(B=\frac{1+3+3^2+...+3^8+3^9}{1+3+3^2+...+3^8}=1+\frac{3^9}{3^8}=4\)
Từ đó suy ra A>B