1 + 2 + 3 + ... + x = 105

=> (1 + x).x:2 = 105

=> (1 + x).x = 105.2

=> (1 + x).x = 210

=> (1 + x).x = 15.14

=> x = 14

Vậy x = 14

\(1+2+3+..+x=105\)

\(\Rightarrow\)\(\frac{\left(x+1\right)x}{2}=105\)

\(\Rightarrow x\left(x+1\right)=210=14\cdot15\)

Vậy x=14

1 + 2 + 3 + ... + x  = 105

(x + 1) . [(x - 1) : 1 + 1] : 2= 105

(x + 1) . [(x - 1) + 1] = 105 . 2

(x + 1) . x = 210

Mà 14 . 15 = 210 => x = 14

Vậy số cần tìm là 14

a: \(35=5\cdot7;105=3\cdot5\cdot7\)

=>\(ƯCLN\left(35;105\right)=5\cdot7=35\)

\(35⋮x;105⋮x\)

=>\(x\inƯC\left(105;35\right)\)

=>\(x\inƯ\left(35\right)\)

=>\(x\in\left\{1;5;7;35\right\}\)

mà x>5

nên \(x\in\left\{7;35\right\}\)

b: \(144=2^4\cdot3^2;192=2^6\cdot3;240=2^4\cdot3\cdot5\)

=>\(ƯCLN\left(144;192;240\right)=2^4\cdot3=48\)

\(144⋮x;192⋮x;240⋮x\)

=>\(x\inƯC\left(192;144;240\right)\)

=>\(x\inƯ\left(48\right)\)

=>\(x\in\left\{1;2;3;4;6;8;12;16;24;48\right\}\)

mà 10<=x<=99

nên \(x\in\left\{12;16;24;48\right\}\)

a,x/9=5/3 = x=15

b,17/x=85/105 = x = 21

c,6/8=15/x = x =  20

d,x/8=2/3 = x = 6

 e,3/x-7=27/135 =x= 22

d) (x – 105) : 21 = 15

(x – 105) = 15. 21

(x – 105) = 315

x = 315+ 105

x = 420

Vậy x = 420

b) x – 105 : 21 = 15

x – 5 = 15

x = 15+5

x = 20

Vậy x = 20

gọi 3 số cần tìm là a,b,c 

ta có 1/2a=1/2b=> a=b

=>>a+b=2a

theo đề ta có:  a+b+c=105

=> 2a+c=105

từ đây ta xét các TH

(Hình như đề bạn ghi thiếu)

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}\)\(+\frac{1}{3}\)

\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

\(\Rightarrow x=93\)

Vậy x=93

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\frac{2.35+8.3}{105}< \frac{x}{105}< \frac{15+42+35}{105}\)

\(94< x< 92\)

Vậy ko có x thỏa mãn

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\Rightarrow\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

\(\Rightarrow x\in\varnothing\)