(2x - 1) x 10 = (2x - 1) x 100
=> 2x - 1 = 0
=> x = 1/2

a) x = 2.               

b) x = 7.

c) x= 12.

d)  x= 45.

e) x = 18.

f) x  = 10.

(2x-1)+(4x-2)+...+(400x-200)=5+10+...+1000

=>\(\left(2x-1\right)+2\left(2x-1\right)+\cdots+200\left(2x-1\right)=5\left(1+2+\cdots+200\right)\)

=>\(\left(2x-1\right)\left(1+2+\cdots+200\right)=5\left(1+2+\cdots+200\right)\)

=>2x-1=5

=>2x=6

=>x=3

2x + 100= 200 + 100

2x + 100 = 300

          2x = 300 - 100

           2x = 200

             x = 100

Vậy x = 100

2x + 100 = 200 + 100

2x + 100 = 300

          2x = 300 - 100

          2x = 200

            x = 100

(2x + 1) + (2x + 2) + (2x + 3) + ... + (2x + 100) = 10050

=> (2x + 2x + .... + 2x) + (1 + 2 + ... + 100) = 10050

=> 100 . 2x + 5050 = 10050

=> 200x = 5000

=> x = 25

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)