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  = \(\frac{5^{202}-1}{24}\)

Trời oi

giúp mik với

a)4⁵*3²+2¹⁰*30=2¹⁰*9+2¹⁰*30=2¹⁰*(9+30)=2¹⁰*39

b) 37-x/3=x-13/7

37+13/7=x+x/3

272/7=4x/3

x=272/7:4/3

x=204/7

\(S=1+5+5^2+5^4+...+5^{200}\)

\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)

\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)

\(\Leftrightarrow25S-S=5^{202}-1\)

\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)

a) S = 1 + 5² + 5⁴ + ... + 5²⁰⁰

=> 5²S = 5².(1 + 5² + 5⁴ + ... + 5²⁰⁰)

=> 25S = 5² + 5⁴ + 5⁶ + ... + 5²⁰²

=> 25S - S = (5² + 5⁴ + 5⁶ + ... + 5²⁰²) - (1 + 5² + 5⁴ + ... + 5²⁰⁰)

=> 24S = 5²⁰² - 1

=> S = \(\frac{5^{202}-1}{24}\)

3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50

a; \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + ... + \(\dfrac{5}{132}\)

 = 5.(\(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ..+ \(\dfrac{1}{132}\))

= 5.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{11.12}\))

= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))

= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))

= 5.(\(\dfrac{6}{12}\) - \(\dfrac{1}{12}\))

= 5.\(\dfrac{5}{12}\)

= \(\dfrac{25}{12}\)

\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)

\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)

 \(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)

  \(A=2^{2011}-2^0\)

\(A=2^{2011}-1\)

\(b,B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)

\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2B=3^{101}-1\)

\(\Rightarrow B=\frac{3^{101}-1}{2}\)

\(c,C=4+4^2+4^3+...+4^n\)

\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)

\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)

\(3C=4^{n+1}-4\)

\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)

\(d,D=1+5+5^2+...+5^{2000}\)

\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)

\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)

\(4D=5^{2001}-1\)

\(\Rightarrow D=\frac{5^{2001}-1}{4}\)

b)

B=1+3+3^2+3^3+..+3^100

=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101

=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)

=> 2B = 3^101 - 1

=> B =( 3^101 - 1) / 2