\(\left(27^{21}-9^{31}-3^{60}\right)\)

\(=\left[\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\right]\)

\(=\left(3^{63}-3^{62}-3^{60}\right)\)

\(=3^{60}\left(3^3-3^2-3\right)\)

\(=3^{60}.17\)

\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)

\(\RightarrowĐPCM\)

\(\left(27^{21}-9^{31}-3^{60}\right)\)

\(=\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\)

\(=\left(3^{63}-3^{62}-3^{60}\right)\)

\(=3^{60}\left(3^3-3^3-3\right)\)

\(=3^{60}.17\)

\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)

Vậy (27²¹ - 9³¹ - 3⁶⁰ ) \(⋮\)17

Ta có:

     \(27^{21}-9^{31}-3^{60}\)

=\(\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\)

=  \(3^{63}-3^{62}-3^{60}\)

=\(3^{60}\left(3^3-3^2-1\right)\)

=\(3^{60}.17⋮17\)

Suy ra: \(27^{21}-9^{31}-3^{60}⋮17\)(đpcm)

\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)

Vậy \(A⋮18\)

\(B=1+3+3^2+...+3^{11}\)

Ta có: \(52=4\cdot13\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)

Vậy \(B⋮4\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)

Vậy \(B⋮13\)

Vì \(4\) và \(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)

Vậy \(B⋮52\)

\(C=3+3^3+3^5+...3^{31}\)

\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)

Vậy \(C⋮15\)

\(D=2+2^2+2^3+...+2^{60}\)

Tao có: \(21=3\cdot7;15=3\cdot5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(D⋮3\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)

Vậy \(D⋮5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có:

\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)

\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)

Vậy \(D⋮15;D⋮21\)

Mình chỉ làm mẫu 1 câu thui nha:

\(A=17^{18}-17^{16}\)

\(A=17^{16}.17^2-17^{16}.1\)

\(A=17^{16}\left(17^2-1\right)\)

\(A=17^{16}.288\)

\(A=17^{16}.16.18\)

\(A⋮18\left(đpcm\right)\)

a)

\(3^{21}-3^{18}\\ =3^{17}.\left(3^4-3\right)\\ =3^{17}.\left(81-3\right)\\ =3^{17}.78\)

Vì \(3^{17}.78⋮78\) nên \(3^{21}-3^{18}⋮78\) (đpcm)

Vậy...

b)
\(81^7-27^9-9^{13}\\ =\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\\ =3^{28}-3^{27}-3^{26}\\ =3^{24}.\left(3^4-3^3-3^2\right)\\ =3^{24}.\left(81-27-9\right)\\ =3^{24}.45\)

Vì \(3^{24}.45⋮45\) nên \(81^7-27^9-9^{13}⋮45\) (đpcm)

Vậy...

đăng từng bài rồi mình giải cho nha

Câu 3,5⁷-5⁶+5⁵=5⁵.5²-5⁵.5+5⁵=5⁵.(5²-5+1)=5⁵.21 chia hết cho 21

Câu:4:7⁶+7⁵-7⁴=7⁴.7²+7⁴.7-7⁴=7⁴.(7²+7-1)=7⁴.55=7⁴.11.5=7³.7.11.5=7³.77.5 chia hết cho 77

Các câu khác tương tự

bạn biết làm hết rồi, chỉ còn câu 2 chưa làm được đúng ko, vậy mình làm cho nhé, nhưng mà mình nghĩ là đề là 81 chứ ko phải 84 đâu

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}=3^{25}\left(3^3-3^2-3\right)=3^{25}.15\) chia hết cho 15

Vậy 81⁷-27⁹-9¹³ chia hết cho 15 (đpcm)

3: \(=5^5\left(5^2-5+1\right)=5^2\cdot21⋮21\)

4: \(=7^4\left(7^2+7-1\right)=7^4\cdot55=7^3\cdot5\cdot77⋮77\)

5: \(=\left(2^{26}+2^{25}-2^{24}\right)=2^{24}\left(2^2+2-1\right)=2^{24}\cdot5⋮5\)

Ta có: \(27^{20}+3^{61}+9^{31}\)

\(=\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\)

\(=3^{60}+3^{61}+3^{62}\)

\(=3^{60}.\left(1+3+3^2\right)\)

\(=3^{60}.13\)

Vì \(13⋮13\) nên \(3^{60}.13⋮13.\)

\(\Rightarrow27^{20}+3^{61}+9^{31}⋮13\left(đpcm\right).\)

Chúc bạn học tốt!