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

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

Chia het cho 31

5^61 + 25^31 + 125^21

= 5^61 + 5^62 + 5^63

= 5^61 x (1+5+25) 

= 5^61 x 31 chia hết 31

5^61 + 25^31 + 125^21

= 5^61 + 5^62 + 5^63

= 5^61 x (1+5+25) 

= 5^61 x 31 chia hết 31

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

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

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

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

\(\Rightarrow A=5^{61}.31⋮31\)

\(\Rightarrow A⋮31\)

Vậy \(A⋮31\)

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

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

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

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

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

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!!!

 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 

giải cụ thể dùng mình nhé

b) \(\left(x-\dfrac{3}{5}\right)^2=4\)

\(\Leftrightarrow\sqrt{\left(x-\dfrac{5}{3}\right)^2}=\sqrt{4}\)

\(\Leftrightarrow\left|x-\dfrac{3}{5}\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=2\\x-\dfrac{3}{5}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)

vậy

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

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

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

\(5^{61}.31⋮31\)

vậy....

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

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

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

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

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

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 .