100x+(1+2+3+...+100)=5050
100x+5050=5050
100x=5050-5050
100x=0
x=0:100
x=0
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Tìm N số tự nhiên x biết
(1+x)+(2+x)+(3+x)+...+(100+x)=5050
BÀI 1:Tìm x biết
a :3x -5 =x
b: 1.2.3.4.5......(x+1)=0
c:x+ 2x +3x +4x +5x+..............+100x=0
d:x +2x+3x+4x+5x+..................+100x=5050
e:(x+1)+(x+2) +(x+3) +(x+4)+............+(x+100)= 5050
Bài 6 Tìm x
1) ( x+1 ) + ( x+2 ) + ( x+3 ) + ... + ( x+100 ) = 5750
2) ( 2x-1 ) + ( 4x-2 ) + ... + ( 200x - 100 ) = 5050
3) ( x+2 ) + ( x+4 ) + ( x+6 ) + ... + ( x + 100) = 2650
Tìm x biết ( 1 + 2 + 3 + 4 + 5 +..............+ 98 + 99 + 100 ) . x = 5050
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5050
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5050
tìm x thuộc Z
a)1+2+3+.........+x=5050
b)1/2+1/6+........1x2+x=99/100
c)1/6+1/12+.......1/x2-x=59/100
d)x-2017+x-2016+.........+99+100=0
g)x-1+x-2+x-3+.......x-2017=0
( x + 1 ) + ( x +2 ) + ( x + 3 ) + .... + ( x + 100 ) = 5050
Câu 1: tìm x, biết!
1+2+3+4+...+x=5050