giúp mình vs mình đang cần gấp T-T

Đặt \(A=1+5+5^2+5^3+...+5^{402}+5^{403}+5^{404}\)

\(\Rightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{399}+5^{400}+5^{401}\right)+\left(5^{402}+5^{403}+5^{404}\right)\)

\(\Rightarrow A=31.1+31.5^3+...+31.5^{402}\)

\(\Rightarrow A=31\left(1+5^3+5^6+...+5^{402}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)

\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)

dpcp là gì vậy ạ

dpcm là j vậy ạ

Hệ số của đơn thức 3x mũ 2y 4xy mũ 3

\(B=5+5^2+5^3+...+5^{88}+5^{89}+5^{90}\)

\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{88}+5^{89}+5^{90}\right)\)

\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{88}\left(1+5+5^2\right)\)

\(=31\left(5+5^4+...+5^{88}\right)⋮31\)

\(B=\left(1+5+5^2\right)+...+5^6\left(1+5+5^2\right)=31\left(1+...+5^6\right)⋮31\)

(3x - 1)³ = 125

(3x - 1)³ = 5³

=>3x - 1 = 5

3x = 5 + 1

3x = 6

x = 6 : 3

x = 2

A = 1+5+5²+5³+...+5⁹⁷+5⁹⁸

A = (1 + 5 + 5²) + (5³ + 5⁴ + 5⁵) + ... + (5⁹⁶ + 5⁹⁷ + 5⁹⁸)

A = 1(1 + 5 + 5²) + 5³(1 + 5 + 5²) + ... + 5⁹⁶(1 + 5 + 5²)

A = 1.31 + 5³.31 + ... + 5⁹⁶.31

A = 31(1 + 5³ + ... + 5⁹⁶)

Vì 31(1 + 5³ + ... + 5⁹⁶) \(⋮\)nên A \(⋮\)31

Vậy A \(⋮\)31

a, \(\left(3x-1\right)^3=125\Leftrightarrow\left(3x-1\right)^3=5^3\)

\(\Rightarrow3x-1=5\Rightarrow3x=5+1\Rightarrow3x=6\Rightarrow x=6\div3=2\)

Vậy x = 2

b, Xét dãy số mũ : 0;1;2;3;...;97;98

Số số hạng của dãy số trên là :

\(\left(98-0\right)\div1+1=99\) ( số )

Ta được số nhóm là :

\(99\div3=33\) ( nhóm )

Ta có : \(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\) (33 nhóm )

\(A=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+...+5^{96}\left(1+5+5^2\right)\)

\(A=1.31+5^3.31+...+5^{96}.31=\left(1+5^3+...+5^{96}\right).31\)

Mà : \(31⋮31;1+5^3+...+5^{96}\in N\Rightarrow A⋮31\) (đpcm)

\(\left(3x-1\right)^3=125\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

Ta có: \(A=1+5+5^2+...+5^{97}+5^{98}\)

\(\Rightarrow A=\left(1+5+5^2\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)

\(\Rightarrow A=31+...+5^{96}\left(1+5+5^2\right)\)

\(\Rightarrow A=31+...+5^{96}.31\)

\(\Rightarrow A=\left(1+...+5^{96}\right).31⋮31\)

Vậy\(A⋮31\)

a, Ta có:

2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100

=  2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100

= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4

=  2 . 31 + 2 6 . 31 + . . . + 2 96 . 31

=  2 + 2 6 + . . . + 2 96 . 31  chia hết cho 31

b, Ta có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5

=  5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6

=  ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6  chia hết cho 6

Ta lại có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150  (có đúng 25 nhóm)

=  [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... +  [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]

=  [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... +  [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]

=  ( 5 . 126 + 5 2 . 126 + 5 3 . 126 ) + ... +  ( 5 145 . 126 + 5 146 . 126 + 5 147 . 126 )

=  ( 5 + 5 2 + 5 3 ) . 126 +  ( 5 7 + 5 8 + 5 9 ) . 126 +  ... + ( 5 145 + 5 146 + 5 147 ) . 126

= 126.[ ( 5 + 5 2 + 5 3 ) + ( 5 7 + 5 8 + 5 9 ) + ... +  ( 5 145 + 5 146 + 5 147 ) ] chia hết cho 126.

Vậy  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150  vừa chia hết cho 6, vừa chia hết cho 126

Chịu 🤭🤭🤭