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)

a) x.31+(1+2+3+4+.....+30)=1240

x.31+465=1240

x.31=1240-465

x.31=775

x=775:31

x=25

b) (1+x).x:2=210

(1+x).x=210.2

(1+x).x=420

Vì 1+x và x là 2 số nhiên liên tiếp và 420=20.21

=> x=20

x-1+x-2+x-3=30

3x-6=30

3x=30+6=36

x=36:3=12

Vậy x=12

\(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)=30\)

\(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)=3.\left(x-2\right)\)

                             \(3.\left(x-2\right)=30\)                     

                              \(3.x-2.3=30\)    

                                 \(3.x-6=30\)             

                                 \(3.x\)        \(=30+6\)

                                 \(3.x\)        \(=36\)

                                    \(x\)        \(=36:3\)

                                    \(x\)        \(=12\)

Vậy \(x=12\)

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=500\)

Số lượng số hạng là: \(30-1+1=30\)

\(\Rightarrow30x+\dfrac{30\left(30+1\right)}{2}=500\)

\(30x+465=500\)

\(30x=35\)

\(x=\dfrac{7}{6}\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=500\)

\(\Rightarrow x+1+x+2+..+x+30=500\)

\(\Rightarrow\left(x+x+...+x\right)+\left(1+2+...+30\right)=500\)

\(\Rightarrow30\cdot x+465=500\)

\(\Rightarrow30\cdot x=500-465\)

\(\Rightarrow30\cdot x=35\)

\(\Rightarrow x=\dfrac{35}{30}\)

\(\Rightarrow x=\dfrac{7}{6}\)

x+(x+1)+(x+2)+........+(x+30)=1246

<=>  31x+31.30:2=1246

<=> 31x+465=1246

<=> 31x=1246-465=781

<=> x=781/31

x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1246

x + x + 1 + x + 2 + ... + x + 30 = 1246

( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1246

Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )

Tổng là : ( 30 + 1 ) x 30 : 2 = 465

=> 31x + 465 = 1246

<=> 31x = 781

<=> x = 781/31

Vậy.......

\(x+(x+1)+(x+2)+(x+3)+....+(x+30)=1240\)

\(=x+\left(30x+1+2+3+...+30\right)=1240\)

\(=31x+1+2+3+...+30\)(có 30 số hạng) \(=1240\)

\(=31x+\left(30+1\right).30\div2=1240\)

\(=31x+465=1240\)

\(\Rightarrow31x=1240-465\)

\(\Rightarrow31x=775\)

\(\Rightarrow x=775\div31\)

\(\Rightarrow x=25\)

đúng cảm ơn nhiều

x+(x+1)+(x+2)+(x+3)+...+(x+30) = 1240

<=> 31x + 465 = 1240 <=> x= 25

\(\text{x+(x+1)+(x+2)+(x+3)+.............+(x+30)=1240}\)

\(=x+x+1+x+2+x+3+...+x+30=1240\)

\(=31x+1+2+3+...+30=1240\)

\(=31x+465=1240\)

\(\Rightarrow31x=1240-465=775\)

\(\Rightarrow x=775:31=25\)

cảm ơn nha mình tk cho bạn bạn tk cho mình nha

\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)

\(\Leftrightarrow x-\dfrac{1}{3}=3\)

\(\Leftrightarrow x=3+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

\(\Leftrightarrow x=\dfrac{3}{20}\)