A>5²⁰¹

Vì khi tính một vài số của A thì đã lớn hơn 5²⁰¹

Ta có:

\(A=5+5^2+5^3+5^4+...+5^{200}\)

\(5A=5.\left(5+5^2+5^3+...+5^{200}\right)\)

\(5A=5^2+5^3+5^4+...+5^{201}\)

\(5A-A=\left(5^2+5^3+5^4+...+5^{200}+5^{201}\right)-\left(5+5^2+5^3+5^4+...+5^{200}\right)\)

\(4A=5^2+5^3+5^4+...+5^{200}+5^{201}-5-5^2-5^3-5^4-...-5^{200}\)

\(4A=\left(5^2-5^2\right)+\left(5^3-5^3\right)+\left(5^4-5^4\right)+...+\left(5^{200}-5^{200}\right)+5^{201}-5\)

\(4A=0+0+0+...+0+5^{201}-5\)

\(4A=5^{201}-5\)

\(A=\frac{5^{201}-5}{4}\)

Vì \(5^{201}-5< 5^{201}\)

\(\Rightarrow\frac{5^{201}-5}{4}< \frac{5^{201}}{4}< 5^{201}\)

hay \(A< 5^{201}\)

Vậy \(A< 5^{201}\)

1, A = 2⁹¹ = 2^7.13 = (2¹³)⁷ = 8192⁷

B = 5³⁵ = 5^5.7 = (5⁵)⁷ = 3125⁷

Vì 3125 < 8192

=> 3125⁷ < 8192⁷

=> B < A

2.Ta có:

 A=11+11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰.

11A=11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰+11²⁰¹.

11A-A=11²⁰¹-11.

10A=11²⁰¹-11.

A=(11²⁰¹-11):10

Quan sát 2 vế A và B thì ta thấy rõ ràng vế A<B hay B>A.

em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

       D =           \(\dfrac{1}{7^2}\) - \(\dfrac{2}{7^3}\) + \(\dfrac{3}{7^4}\) - \(\dfrac{4}{7^5}\) +........+ \(\dfrac{201}{7^{202}}\) -  \(\dfrac{202}{7^{203}}\)

7 \(\times\) D  =  \(\dfrac{1}{7}\) -  \(\dfrac{2}{7^2}\) +  \(\dfrac{3}{7^3}\) - \(\dfrac{4}{7^4}\)  + \(\dfrac{5}{7^5}\) -.......- \(\dfrac{202}{7^{202}}\)

7D +D  =   \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)

         D = (  \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)) : 8

Đặt    B =      \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -........+\(\dfrac{1}{7^{201}}\).-\(\dfrac{1}{7^{202}}\) 

  7   \(\times\) B = 1 - \(\dfrac{1}{7}\)+\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) + \(\dfrac{1}{7^4}\) - \(\dfrac{1}{7^5}\) +.........- \(\dfrac{1}{7^{201}}\)

7B + B   =  1 - \(\dfrac{1}{7^{202}}\)

          B   =  ( 1 - \(\dfrac{1}{7^{202}}\)) : 8

         D  =  [ ( 1 - \(\dfrac{1}{7^{202}}\)): 8  - \(\dfrac{202}{7^{203}}\)] : 8 

          D = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

Từ đầu bài 

=> 5²S=5²+5⁴+5⁶+...+5²⁰²

=>5²S-S= (5²+5⁴+5⁶+...+5²⁰²)-(1+5²+5⁴+...+5²⁰⁰)

=>  24.S = 5²⁰²-1

=>     S  = \(\frac{5^{202}-1}{24}\)

\(A=1+5^2+5^4+.....+5^{2008}\)

\(25A=5^2+5^4+.....+5^{2010}\)

\(25A-A=5^{2010}-1\)

\(A=\frac{5^{2010}-1}{24}<5^{2010}-1\)

`3/(-10) ; 1/(-2) ; 4/(-5)=> -3/10 ; -1/2 ; -4/5`

ta có : `-1/2=(-1xx5)/(2xx5)=-5/10 ; -4/5=(-4xx2)/(5xx2)=-8/10`

vậy `3/(-10) < 1/(-2) < 4/(-5)`

`--------------------`

`2/(-10) ; 7/(-5) ; -1/2=>-2/10 ;-7/5;-1/2`

ta có : `-7/5=(-7xx2)/(5xx2)=-14/10; -1/2=(-1xx5)/(2xx5)=-5/10`

vậy `2/(-10) < -1/2 < 7/(-5)`

`---------------------`

`7/(-4) ; -2/5 ; -3/10=> -7/4;-2/5;-3/10`

ta có : `-7/4=(-7xx5)/(4xx5)=-35/20 ; -2/5=(-2xx4)/(5xx4)=-8/20;-3/10=(-3xx2)/(10xx2)=-6/20`

vậy 7/(-4) > -2/5 > -3/10`

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