CÁC STN LÀ: 0,1,2,3

Ko thể là 4 vì 1246*4=4984

1246 * X < 4972

=> x< 4972:1246

=> x < 3,99...

=> x< 4

=> x\(\in\left\{0;1;2;3\right\}\)

Bài 1 : \(\frac{1}{2}+X.\frac{3}{4}=2\Rightarrow X.\frac{3}{4}=2-\frac{1}{2}=\frac{3}{2}\)

\(\Rightarrow X=\frac{3}{2}:\frac{3}{4}=\frac{3}{2}.\frac{4}{3}=2\)

bài 3: a) =3241+6759+754+1246=10000+2000=12000

b) =9875+125+2141+1859=10000+4000=14000

bài 1 

\(\frac{1}{2}\)+ X x \(\frac{3}{4}=\)2

\(\frac{1}{2}\)+ X = 2 : \(\frac{3}{4}\)

\(\frac{1}{2}\)+ X = \(\frac{8}{3}\)

X = \(\frac{8}{3}\)-- \(\frac{1}{2}\)

X=\(\frac{13}{6}\)

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

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

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

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

a)

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

x + x + 1 + x + 2 + ... + x + 30 = 1240

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

(x . [30 - 1 + 1 + 1]) + ([30 + 1] . [30 - 1 + 1] : 2) = 1240

31x + 465 = 1240

           31x = 1240 - 465

           31x = 775

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

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

\(31x+465=1240\)

\(31x=775\)

\(x=25\)

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

=>x.31+465=1240

=>x.31=1240-465

=>x.31=775

=>x=775:31

=>x=25

vậy x=25

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25 nha

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