mk ko bt 123

Ta có:

\(24^{54}.54^{24}.2^{10}=\left(2^3.3\right)^{54}.\left(3^3.2\right)^{24}.2^{10}\)

\(=\left(2^3\right)^{54}.\left(3^3.2\right)^{24}.2^{10}\)

\(=2^{162}.3^{54}.3^{72}.2^{24}.2^{10}\)

\(=2^{196}.3^{126}\) (1)

Lại có:

\(72^{63}=\left(2^3.3^2\right)^{63}=2^{189}.3^{126}\)(2)

Từ (1) và (2) ⇒ \(24^{54}.54^{24}.2^{10}⋮72^{63}\)

Bấm vào đây /hoi-dap/question/132960.htm

Vào câu trả lời tương tự

3^n+2 - 2^n+2 + 3^n - 2^n = (3ⁿ⁺²+3ⁿ)+(-2ⁿ⁺²-2ⁿ)

=3ⁿ.(3²+1)-2ⁿ.(2²+1)

=3ⁿ.10-2ⁿ.5

=3ⁿ.10-2^n-1.2.5

=3ⁿ.10-2^n-1.10

=10.(3ⁿ-2^n-1)

Vậy 3^n+2 - 2^n+2 + 3^n - 2^n chia hết cho 10

a, \(81^7-27^9-9^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=3^{22}\left(3^6-3^5-3^4\right)\)

\(=3^{22}\times405⋮405\)

\(2^{54}.54^{24}.2^{10}\)chia hết \(72^{63} \)

\(2^{54}.54^{24}.2^{10}\)=\((2^3.3)^{54}.(3^3.2)^{24}.2^{10}\)

=\((2^3)^{54}.3^{54}.(3^3)^{24}.2^{24}2^{10}\)

= \(2^{162}.2^{24}.2^{10}.3^{54}.3^{72} \)

=\(2^{196}.3^{126}\)

\(72^{63} \)=\((2^3.3^2)^{63}\)

=\((2^3)^{63}.(3^2)^{63}=2^{189}.3^{126}\)

Vì \(2^{196}.3^{126}\)chia hết \(2^{189}.3^{126}\)\(24^{54}.54^{24}.2^{10}\)

\(\Rightarrow \)\(24^{54}.54^{24}.2^{10}\)chia hết \(72^{63} \)(dpcm)

24^54 * 54^24 * 2^10 = (3 * 2^3)^54 * (2 * 3^3)^24 * 2^10 
= 3^126 * 2^196(1)
72^63 = (2^3 * 3^2) ^63 = 2^189 * 3^126(2)

từ (1) và (2)=>24^54 * 54^24 * 2^10 chia hết 72^63

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)

=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\frac{97}{198}\)

=\(\frac{-95}{198}\)