Từ đầu bài 

=> 5²S=5²+5⁴+5⁶+...+5²⁰²

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

=>  24.S = 5²⁰²-1

=>     S  = \(\frac{5^{202}-1}{24}\)

Lời giải:
Gọi tổng trên là $K$
$K=1+5^2+5^3+5^4+...+5^{200}$

$5K=5+5^3+5^4+5^5+...+5^{201}$
$\Rightarrow 5K-K = 5+5^{201}-1-5^2$

$\Rightarrow 4K = 5^{201}-21$

$\Rightarrow K= \frac{5^{201}-21}{4}$

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)

chán!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a.S=1+5²+5⁴+...+5²⁰⁰

=>25S=5²+5⁴+5⁶+...+5²⁰²

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

=>24S=5²⁰²-1

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

b.ta có:

\(\frac{a-1}{2}=\frac{5a-5}{10};\frac{b+3}{4}=\frac{3b+9}{12};\frac{c-5}{6}=\frac{4c-20}{24}\)

\(\Rightarrow\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}=\frac{5a-5-3b-9-4c+20}{10-12-24}=\frac{\left(5a-3b-4c\right)+\left(20-9-5\right)}{-26}\)

\(=\frac{46+6}{-26}=\frac{52}{-26}=-2\)

\(\Rightarrow a-1=-2.2=-4\Rightarrow a=-3\)

\(\Rightarrow b+3=-2.4\Rightarrow b=-11\)

\(\Rightarrow c-5=-2.6=-12\Rightarrow c=-7\)

vậy a=-3;b=-11;c=-7

\(\frac{a-1}{2}\) = \(\frac{b+3}{4}\)=\(\frac{c-5}{6}\)và 5a-3b-4c=46

\(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}=k\)\(\overline{1}\)

a=2k+1 

b= 4k-3

c=6k+5

Thay vào \(\overline{1}\)ta đc : 5(2k+1)-3(4k-3)-4(6k+5)=46

=10k+5-12k-9-32k+20=46

=\(\frac{10k-32k-12k}{5-9-20}=-\frac{46}{24}=-\frac{23}{12}\)??????????????????

Lời giải:

\(S=1+5^2+5^4+....+5^{198}+5^{200}\) (1)

\(\Rightarrow 5^2.S=5^2+5^4+...+5^{200}+5^{202}\) (2)

Lấy (2) trừ (1):

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

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

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

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

\(5^2S=5^2+5^4+5^6+...+5^{202}.\)

\(5^2S-S=\left(5^2+5^4+5^6+...+5^{202}\right)-\left(1+5^2+5^4+...+5^{200}\right).\)

\(24S=5^{202}-1\Rightarrow S=\dfrac{5^{202}-1}{24}.\)

Vậy.....

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

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

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

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

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

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 = \(\frac{5^{202}-1}{24}\)