\(5^{61}+25^{31}+125^{21}=5^{61}+5^{62}+5^{63}\)

                                         \(=5^{61}\left(1+5+5^2\right)=5^{61}.31\)

Chia het cho 31

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)

\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1}{2009}\)

1,

\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)

\(|x- \frac{2}{7}|=\frac{1}{7}\)

<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)

Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)

Học tốt

\(5^{61}+25^{31}+125^{21}\)

\(=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)

\(=5^{61}\cdot5^{2\cdot31}\cdot5^{3\cdot21}\)

\(=5^{61}+5^{62}+5^{63}\)

\(=5^{61}\cdot\left(1+5+5^2\right)\)

\(=5^{61}\cdot\left(6+5^2\right)\)

\(=5^{61}\cdot\left(6+25\right)\)

\(=5^{61}\cdot31\)

Vì \(5^{61}\inℤ\)

\(\Rightarrow5^{61}\cdot31⋮31\)

\(\Rightarrow5^{61}+25^{31}+125^{21}⋮31\)

Vậy bài toán đã được chứng minh . 

5^61+25^31+125^21 =5^61+5^62+5^63 =5^61(1+5+5^2) =5^61.31

Không chia hết cho 3 đâu bạn, chỉ 31 thôi

a ) \(5^{61}+25^{31}+125^{21}=5^{61}+5^{62}+5^{63}=5^{61}\left(1+5+25\right)=5^{61}.31⋮31\)(đpcm)

b ) \(6^3+2.6^2+3^3=2^3.3^3+2^3.3^2+3^3=3^2\left(8.3+8+3\right)=3^2.35⋮35\) (đpcm)

Vậy ........

Cảm ơn các bạn nhiều lắm nha!!!

\(A=5^{61}+25^{31}+125^{21}\)

\(\Rightarrow A=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)

\(\Rightarrow A=5^{61}+5^{62}+5^{63}\)

\(\Rightarrow A=5^{61}\left(1+5+5^2\right)\)

\(\Rightarrow A=5^{61}.31⋮31\)

\(\Rightarrow A⋮31\)

Vậy \(A⋮31\)

\(A=5^{61}+25^{31}+125^{21}\)

\(A=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)

\(A=5^{61}+5^{62}+5^{63}\)

\(A=5^{61}\left(1+5+5^2\right)\)

\(A=5^{61}\cdot31⋮31\left(đpcm\right)\)

 5⁶¹ + 25³¹ + 125²¹ = 5⁶¹ + (5²)³¹ + (5³)²¹ = 5⁶¹ + 5⁶² + 5⁶³ = 5⁶¹ + 5⁶¹ . 5 + 5⁶¹ . 5² = 5⁶¹(1 + 5 + 5²) 

= 5⁶¹ . 31

có: 155 = 31 . 5

=> 5⁶¹ . 31 chia hết cho 31 . 5 

\(=5^{20}+\left(5^2\right)^{11}+\left(5^{ }^3\right)^7\)

=\(5^{^{ }20}+5^{22}+5^{21}\)

\(=5^{20}\cdot\left(1+5^2+5^1\right)\)

=\(5^{20}\cdot\left(1+25+5\right)\)

=\(5^{20}\cdot31\)

Vì 31 chia hết chó 31 nên

\(5^{20}+25^{^{ }11}+125^7\)chia hết cho 31

\(^{5^{20}+25^{11}+125^7}\)=\(1.5^{20}+25.25^{10}+\left(5^3\right)^7\)=\(1.5^{20}+25.\left(5^2\right)^{10}+5^{21}\)=\(1.5^{20}+25.5^{20}+5.5^{20}\)

=\(^{5^{20}.\left(1+25+5\right)}\)=\(5^{20}.31\)chia hết cho 31

Vậy \(5^{20}+25^{11}+125^7\)chia hết cho 31

5^20+25^11+125^7=5^20+(5^2)^11+(5^3)^7= 5^20+5^22+5^21=5^20(1+5^2+5)=5^20.31

Vậy 5^20+25^11+125^7 chia hết cho 31

giải cụ thể dùng mình nhé

ta có(^ là dấu mũ):

5^20+25^11+125^7=5^20+5^22+5^21

=5^20+5^20.5^2+5^21.5

=5^20.(1+5^2+5)=5^20.(1+25+5)=5^20.31 chia hết cho 31

Nếu sai chỗ nào thì nhắc mik nhé :)

\(5^{20}+25^{11}+125^7=5^{20}+5^{2^{11}}+5^{3^7}=5^{20}+5^{22}+5^{21}=5^{20}+5^{20}.5^2+5^{20}.5=5^{20}\left(5^2+5+1\right)=5^{20}.31\)Vì \(5^{20}.31⋮31\) nên \(\left(5^{20}+25^{11}+125^7\right)⋮31\)