bằng 2000 nha bạn 

k nha

\(2000\div5\times5=2000\)

Ai k mk mk k lại ! 

2000:5x5=2000

Dep trai la phai the. 

200:5x5=40x5=200

k mình đi bạn

\(200\div5\times5=200\)\

Ai k mình mình k lại ! 

200 : 5 x 5 = 200

=78125

78125

k cho mk nhé

`24 xx 5^5 + 5^2 xx 5^3`

`=24 xx 5^5 + 5^5`

`=5^5 (24 +1)`

`=5^5 . 25`

`=5^5 .5^2`

`=5^7`

=24x3125 +\(5^2+5^3\)

=75000+ \(5^5\)

=75000+3125

=78125

\(24\cdot5^5+5^2\cdot5^3\)

\(=24\cdot5^5+5^5\)

\(=5^5\cdot5^2=5^7\)

Ta có: Phân số \(\dfrac{0}{7}\) có tử số là 0, mẫu số là 7

Phân số \(\dfrac{3}{-8}\) có tử số là 3, mẫu số là -8

\(-29\times165+29\times65=29\times\left(65-165\right)=29\times\left(-100\right)=-2900\\ ---\\ \left\{215-\left[5\times\left(5\times16-25\times2\right)\right]-60\right\}:5^3\\ =\left\{215-\left[5\times\left(80-50\right)\right]-60\right\}:125\\ =\left\{215-\left[5\times30\right]-60\right\}:125\\ =\left\{215-150-60\right\}:125=5:125=\dfrac{1}{25}\)

Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)

=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)

=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)

=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)

=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)

Ta có:

\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)

\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)

\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)

\(\left(125^3\cdot7^5-175^5:5\right):2001^{2002}\)

\(\left(5^9\cdot7^5-7^5\cdot5^{10}:5\right):2001^{2002}\)

\(\left(5^9\cdot\left(7^5-7^5\right)\right):2001^{2002}\)

\(\left(5^9\cdot0\right):2001^{2002}\)

\(0:2001^{2002}\)

= \(0\)