\(=\dfrac{5^4\cdot5^4\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{5^2}\cdot\dfrac{1}{2^2}=\dfrac{1}{100}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.4^4.5^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^8.5^2.4^4.4}=\frac{1}{25.4}=\frac{1}{100}\)

a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)

b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)

c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)

\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)

\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)

\(\Rightarrow C=1-\frac{1}{2^{100}}\)

d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)

\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)

\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)

\(\Rightarrow4D=5-\frac{1}{5^{101}}\)

\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)

a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)

b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)

c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)

\(C=1-\frac{1}{2^{100}}\)

phần d bn lm tương tự như phần c nha!
 

Có bn nào lm giùm mik cách khác cho câu D được không?

-6/35

1/5x (4/7- 3/7- 1)=1/5x -6/7= -6/35

\(-\dfrac{6}{35}\)

\(\dfrac{1}{5}\times\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}-\dfrac{1}{5}\)

\(=\dfrac{4}{35}+\dfrac{3}{35}-\dfrac{1}{5}\)

\(=\dfrac{7}{35}-\dfrac{1}{5}\)

\(=\dfrac{1}{5}-\dfrac{1}{5}\)

\(=0\)

0

`a)` `2/5 + 3/5 xx 4/9 - 1/3`

`= 2/5 + 4/15 - 1/3`

`= 6/15 + 4/15 - 5/15`

`= 1/3`

`b)` `( 5/2 - 1/3 ) xx 9/2 - 1/2`

`= ( 15/6 - 2/6 ) xx 9/2 - 1/2`

`= 13/6 xx 9/2 - 1/2`

`= 39/4 - 2/4`

`= 37/4`

5 x 3 + 5 x 4 + 5 x 2 + 5

= 5 x 3 + 5 x 4 + 5 x 2 + 5 x 1

= 5 x ( 3 + 4 + 2 + 1 )

= 5 x 10

= 50

  5 x 3 + 5 x 4 + 5 x 2 + 5

 = 5 x (3 + 4 + 2) + 5

= 5 x 9 + 5

= 45 + 5

= 50

=5x(3+4+2+1)

=5x10

=50

1-3/5

=5/5-3/5

=2/5

3/5*4/7

=12/35

2/5 >12/35

nho ****