Question

Tính A= 5^28+5^24+...+5^4+1/5^30+5^28+5^26+...+5^2+1

Answer 1

Mình cần gấp. Mong các bạn giúp đỡ

Answer 2

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 \(C=5^{30}+5^{28}+5^{26}+........+5^2+1\)

\(\Rightarrow5^2.C=5^{32}+5^{30}+5^{28}+.........+5^4+5^2\)

\(\Rightarrow25.C=5^{32}+5^{30}+5^{28}+........+5^4+5^2\)

\(\Rightarrow25.C-C=24.C=5^{32}-1\)

\(\Rightarrow C=\frac{5^{32}-1}{24}\)

\(\Rightarrow A=\frac{B}{C}=\frac{5^{32}-1}{624}:\frac{5^{32}-1}{24}=\frac{5^{32}-1}{624}.\frac{24}{5^{32}-1}=\frac{24}{624}=\frac{1}{26}\)

Answer 3

Đặt \(B=5^{28}+5^{24}+...+5^4+1\Rightarrow5^4B=5^{32}+5^{28}+...+5^8+5^4\)

\(\Rightarrow\left(5^4-1\right)B=5^{32}-1\Rightarrow B=\frac{5^{32}-1}{5^4-1}\)

Đặt \(C=5^{30}+5^{28}+...+5^2+1\Rightarrow5^2C=5^{32}+5^{30}+...+5^4+5^2\)

\(\Rightarrow\left(5^2-1\right)C=5^{32}-1\Rightarrow C=\frac{5^{32}-1}{5^2-1}\)

Khi đó \(A=\frac{B}{C}=\frac{\frac{5^{32}-1}{5^4-1}}{\frac{5^{32}-1}{5^2-1}}=\frac{5^2-1}{5^4-1}=\frac{1}{26}\)

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