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?

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

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

.

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:  \(B=\frac{7^{32}-1}{7^2-1}\)

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

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

=\(=\frac{7^{32}-1}{7^{32}-1}\)

=1

vậy C=1

lu nhu ngu

Bài làm

\(A=\frac{2^2.10+2^3.6}{2^2.15-2^4}\)

\(A=\frac{2^2.10+2.2^2.6}{2^2.15-2^2.2^2.1}\)

\(A=\frac{2^2.\left(10+6\right).2}{2^2.\left(15-1\right).2^2}\)

\(A=\frac{2^2.16.2}{2^2.14.2^2}\)

\(A=\frac{16}{14.2}\)

\(A=\frac{8}{7.2}\)

\(A=\frac{8}{14}\)

\(A=\frac{4}{7}\)

Vậy \(A=\frac{4}{7}\)

\(B=\frac{2^9.15^{17}.75^3}{18^8.5^{24}.9^2}\)

\(B=\frac{2^9.\left(3.5\right)^{17}.\left(3.5^2\right)^3}{\left(2.3^2\right)^8.5^{24}.\left(3^2\right)^2}\)

\(B=\frac{2^9.3^{17}.5^{17}.3^3.5^6}{2.3^{19}.5^{24}.3^4}\)

\(B=\frac{2^8.1.1.1.5}{1.3^2.1.3}\)

\(B=\frac{2^8.5}{3^3}\)

\(B=\frac{1280}{27}\)

mình nghi thiếu dấu cộng sorry các bạn

Học sinh nhẩm và ghi kết quả như sau:

35 : 5 = 7

28 : 7 = 4

24 : 6 = 4

21 : 3 = 7

35 : 7 = 5

28 : 4 = 7

24 : 4 = 6

21 : 7 = 3

35 : 5 = 7

28 : 7 = 4

24 : 6 = 4

21 : 3 = 7

35 : 7 = 5

28 : 4 = 7

24 : 4 = 6

21 : 7 = 3