Question

Tính

\(C=\frac{7^{28}+7^{24}+...+7^4+7^0}{7^{30}+7^{28}+...+7^2+7^0}\)

Ai nhanh 3 tick !

Answer 1

Cách tui lẹ hơn cách bạn Nguyễn Duy Khánh nè!

Ta có: \(C=\frac{7^{28}+7^{24}+....+7^4+7^0}{\left(7^{30}+7^{26}+...+7^6+7^2\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)

\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{7^2\left(7^{28}+7^{24}+...+7^4+7^0\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)

\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{\left(7^{28}+7^{24}+...+7^4+7^0\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)

P/s: Easy đúng không?

Answer 2

\(C=\frac{7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0}{7^{30}+7^{28}+...+7^2+7^0}\)

Đặt A là tử số ,B là mẫu số.Ta có:

\(7^{\text{4}}A=7^{32}+7^{28}+...+7^8+7^{\text{4}}+7^0\)

\(20\text{4}1A-A=\left(7^{32}+7^{28}+7^{2\text{4}}+...+7^8+7^{\text{4}}\right)-\left(7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0\right)\)

\(2\text{4}00A=7^{32}-7^0=7^{32}-1\)

\(\Rightarrow A=\left(7^{32}-1\right):2\text{4}00\)

\(7^2B=\left(7^{32}+7^{30}+7^{28}+...+7^{\text{4}}+7^2\right)\)

49B-B= ....tự..điền......như A nhé.....

48B=7³²-1 =>B=[72³²-1]:48

=>\(C=\frac{A}{B}=\frac{\left(7^{32}-1\right):2\text{4}00}{\left(7^{32}-1\right):\text{4}8}\)

Tui nghĩ vậy đc r á

p/s:ko chắc

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Answer 3

Đặt:  \(A=7^{28}+7^{24}+...+7^4+7^0\)

=>  \(7^4.A=7^{32}+7^{28}+...+7^8+7^4\)

=>  \(7^4.A-A=\left(7^{32}+7^{28}+...+7^8+7^4\right)-\left(7^{28}+7^{24}+...+7^4+7^0\right)\)

=>  \(\left(7^4-1\right).A=7^{32}-1\)

=>  \(A=\frac{7^{32}-1}{7^4-1}\)

Đặt:  \(B=7^{30}+7^{28}+...+7^2+7^0\)

tương tự tính được:  \(B=\frac{7^{32}-1}{7^2-1}\)

\(C=\frac{A}{B}=\frac{7^{32}-1}{7^4-1}:\frac{7^{32}-1}{7^2-1}=\frac{7^2-1}{7^4-1}\)

\(=\frac{7^2-1}{\left(7^2-1\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)