Giải

Ta có : 5.A = 5 . ( 1 + 5 + 5² + .... + 5²⁰¹⁹ )

                  = 5 + 5² +5 ³ + ... + 5²⁰²⁰

Suy ra : 5A - A = ( 5 + 5² + 5³ + ... + 5²⁰¹⁹ + 5²⁰²⁰ ) - ( 1 + 5 + 5² + ... + 5²⁰¹⁹ )

              A = 5²⁰²⁰ - 1

Vậy A = 5²⁰²⁰ -1

# Chúc bạn học tốt #

5A = 1 + 5.( 5 + 5² + 5^x + .... + 5²⁰¹⁹ )

5A = 1 + 5² + 5^x +....+ 5²⁰²⁰

5A - A = 1 + ( 5² + 5^x +.....+ 5²⁰²⁰ ) - ( 5 + 5² + 5^x +.....+ 5²⁰¹⁹ )

4A = 1 + 5²⁰²⁰ - 5

A = 1 + ( 5²⁰²⁰ - 5 ) : 4

Sai thì thôi bn nhé !!!

10² = 10

10³ = 1 000

10⁴ = 10 000

10⁵ = 100 000

10⁶ = 1 000 000

10\(^2\)=100

10\(^3\)=1000

10\(^4\)=10000

10\(^5\)=100000

10\(^6\)=1000000

K nhé ????

chịu ??????

em mới học lớp 5 thôi mà

cả nhà giúp mik nhá .............. 

chịu hẳn , bạn ghi rõ ra ..

Chị Linh giúp e với ạ!

   234 - ( 56 + 24 : 4 )
=234 - ( 56 + 6 )
=234 - 62
=172
 

  100 : { 250 : [ 450 - ( 500 - 2² . 25) ] }
=100 : { 250 : [ 450 - ( 500 - 4 . 25) ] }
=100 : { 250 : [ 450 - ( 500 - 100) ] }
=100 : { 250 :  [ 450 - 400] }
=100 : { 250 : 50}
=100 : 5
=20

a: \(\frac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(=\frac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}=\frac{2^8\cdot5^7\left(2^3+3^3\right)}{2^5\cdot5^4\left(2^3+3^3\right)}\)

\(=2^3\cdot5^3=10^3=1000\)

b: \(149-\left(35:x+3\right)\cdot17=13\)

=>\(\left(35:x+3\right)\cdot17=149-13=136\)

=>35:x+3=136:17=8

=>35:x=8-3=5

=>\(x=\frac{35}{5}=7\)

c: \(121:11-\left(4x+5:3\right)=4\)

=>11-(4x+5/3)=4

=>4x+5/3=11-4=7

=>\(4x=7-\frac53=\frac{21}{3}-\frac53=\frac{16}{3}\)

=>\(x=\frac{16}{3}:4=\frac43\)

d: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=40\)

=>\(41-\left(2x-5\right)=\frac{720}{40}=18\)

=>2x-5=41-18=23

=>2x=28

=>\(x=\frac{28}{2}=14\)

e: Sửa đề: (x+1)+(x+2)+(x+3)+...+(x+100)=5700

=>100x+(1+2+3+...+100)=5700

=>\(100x+100\cdot\frac{101}{2}=5700\)

=>x+50,5=57

=>x=57-50,5

=>x=6,5

a)\(x^{15}=x\Rightarrow x\in\left\{0;1\right\}\)                                            d)Là ý b

b)\(\left(2x+1\right)^3=125\\ \left(2x+1\right)^3=5^3\Rightarrow2x+1=5\\ 2x=4\\ x=2\)                       e)\(\left(x-3\right)^2=25\\ \Rightarrow\left(x-3\right)^2=5^2\\ \Rightarrow\hept{\begin{cases}x-3=5\Rightarrow x=8\\x-3=-5\Rightarrow x=-2\end{cases}}\)

c)\(\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow x-5\in\left\{0;1\right\}\Rightarrow x\in\left\{5;6\right\}\)

\(\left(x+1\right)^3=27\)

\(\left(x+1\right)^3=3^3\)

\(\Rightarrow x+1=3\)

\(x=2\)

\(\left(x+1\right)^3=27\)

\(< =>\left(x+1\right)^3=3.3.3=3^3\)

\(< =>x+1=3< =>x=3-1=2\)

\(\left(2x+3\right)^3=9.81\)

\(< =>\left(2x+3\right)^3=9.9.9\)

\(< =>\left(2x+3\right)^3=9^3\)

\(< =>2x+3=9< =>2x=6\)

\(< =>x=\frac{6}{2}=3\)