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A= 1+5^1
A x5=5+5^2+5^3+5^4+5^5+5^6+5^7+.......5^19+5^20
Ax5-A=5^20-1
A=(5^20-1):4
Vậy A=(5^20-1):4.Nhớ k đó
Ta có :
\(A=\frac{5^5+2}{5^5-1}=\frac{5^5-1}{5^5-1}+\frac{3}{5^5-1}\)
\(=1+\frac{3}{5^5-1}\)
\(B=\frac{5^5}{5^5-3}=\frac{5^5-3}{5^5-3}+\frac{3}{5^5-3}\)
\(=1+\frac{3}{5^5-3}\)
\(5^5-1>5^5-3\)
\(\Rightarrow\frac{3}{5^5-1}< \frac{3}{5^5-3}\)
\(\Rightarrow1+\frac{3}{3^5-1}< 1+\frac{3}{3^5-3}\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
\(A=\frac{5^5+2}{5^5-2}>\frac{5^5}{5^5-1}>\frac{5^5}{5^5-3}=B\Rightarrow A>B\)
lý Kì Anh Bạn nhầm rồi nhé :)
\(5^5-1>5^5-3\)nên \(\frac{5^5}{5^5-1}< \frac{5^5}{5^5-3}\)
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
Đặt \(A=\frac{5^{28}+5^{24}+.........+5^4+1}{5^{30}+5^{28}+5^{26}+.....+5^2+1}=\frac{B}{C}\)
Xét tử \(B=5^{28}+5^{24}+........+5^4+1\)
\(\Rightarrow5^4.B=5^{32}+5^{28}+........+5^8+5^4\)
\(\Rightarrow625.B=5^{32}+5^{28}+........+5^8+5^4\)
\(\Rightarrow625.B-B=624.B=5^{32}-1\)
\(\Rightarrow B=\frac{5^{32}-1}{624}\)
Xét mẫu \(C=5^{30}+5^{28}+5^{26}+........+5^2+1\)
\(\Rightarrow5^2.C=5^{32}+5^{30}+5^{28}+.........+5^4+5^2\)
\(\Rightarrow25.C=5^{32}+5^{30}+5^{28}+........+5^4+5^2\)
\(\Rightarrow25.C-C=24.C=5^{32}-1\)
\(\Rightarrow C=\frac{5^{32}-1}{24}\)
\(\Rightarrow A=\frac{B}{C}=\frac{5^{32}-1}{624}:\frac{5^{32}-1}{24}=\frac{5^{32}-1}{624}.\frac{24}{5^{32}-1}=\frac{24}{624}=\frac{1}{26}\)
???????????
hahahaha