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Số phần tử chia hết cho 4 là: \(4,0,12,8\)

Vậy có 4 phần tử chia hết cho 4 trong tập hợp A 

Tập hợp các phần tử chia hết cho 4 A={ 0 ; 4 ; 12 ; 8 }

Vậy có 4 phần tử chia hết cho 4

\(96=2\cdot48=2\cdot2\cdot24=2\cdot2\cdot2\cdot2\cdot2\cdot3=2^5\cdot3\)

\(127=127\) (do 127 là số nguyên tố)

\(260=2\cdot130=2\cdot10\cdot13=2\cdot2\cdot5\cdot13=2^2\cdot5\cdot13\)

\(312=2\cdot156=2\cdot2\cdot78=2\cdot2\cdot2\cdot39=2^3\cdot3\cdot13\)

\(860=2\cdot430=2\cdot2\cdot215=2^2\cdot5\cdot43\)

\(4\left(x-5\right)-16=24\cdot3\)

\(\Rightarrow4\left(x-5\right)-16=72\)

\(\Rightarrow4\left(x-5\right)=72+16\)

\(\Rightarrow4\left(x-5\right)=88\)

\(\Rightarrow x-5=\dfrac{88}{4}\)

\(\Rightarrow x-5=22\)

\(\Rightarrow x=22+5\)

\(\Rightarrow x=27\)

\(4.\left(x-5\right)-16=24.3\)

\(4.\left(x-5\right)-16=72\)

\(4.\left(x-5\right)=72+16\)

\(4.x-5=88\)

\(x-5=88:4\)

\(x-5=22\)

\(x=22+5\)

\(x=27\)

`#3107.101107`

A,

\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)

B,

\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)

Help

`#3107.101107`

Gọi biểu thức trên là A

Ta có:

\(A=1+5^2+5^4+...+5^{40}\\ =1\cdot\left(1+5^2\right)+5^4\cdot\left(1+5^2\right)+...+5^{38}\cdot\left(1+5^2\right)\\ =\left(1+5^2\right)\cdot\left(1+5^4+...+5^{38}\right)\\ =26\cdot\left(1+5^4+...+5^{38}\right)\)

Vì \(26\cdot\left(1+5^4+...+5^{38}\right)\text{ }⋮\text{ }26\)

\(\Rightarrow A\text{ }⋮\text{ }26\)

_______

Gọi biểu thức trên là B

Ta có:

\(B=1+2^2+2^4+...+2^{100}\\ =1\cdot\left(1+2^2+2^4\right)+2^6\cdot\left(1+2^2+2^4\right)+...+2^{96}\cdot\left(1+2^2+2^4\right)\\ =\left(1+2^2+2^4\right)\cdot\left(1+2^6+...+2^{96}\right)\\ =21\cdot\left(1+2^6+...+2^{96}\right)\)

Vì \(21\cdot\left(1+2^6+...+2^{96}\right)\text{ }⋮\text{ }21\)

\(\Rightarrow B\text{ }⋮\text{ }21\)

_______

Gọi biểu thức trên là C

Ta có:

\(C=1+3^2+3^4+...+3^{100}\\ =1\cdot\left(1+3^2+3^4+3^6\right)+3^6\cdot\left(1+3^2+3^4+3^6\right)+...+3^{94}\cdot\left(1+3^2+3^4+3^6\right)\\ =\left(1+3^2+3^4+3^6\right)\cdot\left(1+3^6+...+3^{94}\right)\\ =820\cdot\left(1+3^6+...+3^{94}\right)\)

Vì \(820\cdot\left(1+3^6+...+3^{94}\right)\text{ }⋮\text{ }82\)

\(\Rightarrow C\text{ }⋮\text{ }82.\)

a) \(A=1+5^2+5^4+5^6...+5^{40}\)

\(\Rightarrow A=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{38}\left(1+5^2\right)\)

\(\Rightarrow A=26+5^4.26+...+5^{38}.26\)

\(\Rightarrow A=26\left(1+5^4+...+5^{38}\right)⋮26\)

\(\Rightarrow1+5^2+5^4+5^6...+5^{40}⋮6\left(dpcm\right)\)

b) \(B=1+2^2+2^4+2^6+...+2^{100}\)

\(\Rightarrow B=\left(1+2^2+2^4\right)+2^6\left(1+2^2+2^4\right)+...+2^{96}\left(1+2^2+2^4\right)\)

\(\Rightarrow B=21+2^6.21+...+2^{96}.21\)

\(\Rightarrow B=21\left(1+2^6+...+2^{96}\right)⋮21\)

\(\Rightarrow1+2^2+2^4+2^6+...+2^{100}⋮21\left(dpcm\right)\)

Bài C tương tự bạn tự làm nhé!