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ta có
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)\(-4\)
\(\Rightarrow\left(\frac{x+1}{212}+1\right)+\left(\frac{x+2}{211}+1\right)+\left(\frac{x+3}{210}+1\right)+\left(\frac{x+4}{209}+1\right)=-4+4\)
=> \(\frac{x+1+212}{212}+\frac{x+2+211}{211}+\frac{x+3+210}{210}+\frac{x+4+209}{209}\) =\(0\)
=> \(\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}\)=\(0\)
=> (x+213) \(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)=0
mà\(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)\(\ne0\)
=>x+213=0 => x=-213
vậy x= -213
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)
\(\Rightarrow\frac{x+1}{212}+1+\frac{x+2}{211}+1+\frac{x+3}{210}+1+\frac{x+4}{209}+1=-4+4=0\)
\(\Rightarrow\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}=0\)
\(\Rightarrow\left(x+213\right)\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)=0\)
\(\Rightarrow x+213=0\Leftrightarrow x=-213\)

Ta có :
\(\frac{1}{3}+\frac{3}{35}=\frac{70}{210}+\frac{18}{210}=\frac{88}{210}\)
\(\frac{4}{7}+\frac{3}{5}+\frac{1}{3}=\frac{120}{210}+\frac{126}{210}+\frac{70}{210}=\frac{316}{210}\)
⇒ 88 < x < 316

a: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)

a,Ư(25)={1;5;25}
\(\Rightarrow\)2x+1\(\in\left\{1;5;25\right\}\)
\(\Rightarrow x\in\left\{0;2;12\right\}\)
bTa có : \(x\inƯC\left(60;150;210\right)\)
Ư(60)={1;2;3;4;5;6;10;12;15;20;30;60}
Ư(150)={1;2;3;5;6;10;15;25;30;50;125;150}
Ư(210)={1;2;3;5;6;7;10;15;30;35;42;70;105;210}
\(\Rightarrow x\in\left\{1;2;3;5;6;10;15;30\right\}\)
mà x>25 nên x=30

a) 169 . ( 3x - 9.17 ) + 24 : 3 = 30
169 . ( 3x - 153 ) + 8 = 30
169 . ( 3x - 153 ) = 30 - 8
169 . ( 3x - 153 ) = 22
3x - 153 = 22 : 169
3x - 153 = \(\frac{22}{169}\)
3x = \(\frac{22}{169}+153\)
3x = \(\frac{25879}{169}\)
x = \(\frac{25879}{169}:3\)
x = \(\frac{25879}{507}\)
Vậy \(x=\frac{25879}{507}\)
b) \(\left(\frac{4}{5}:\frac{6}{5}+\frac{1}{5}:\frac{1}{x}\right).30-26=54\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=54+26\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=80\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right)=80:30\)
\(\frac{2}{3}+\frac{1}{5}.x=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}-\frac{2}{3}\)
\(\frac{1}{5}.x=2\)
\(x=2:\frac{1}{5}\)
\(x=10\)
Vậy \(x=10\)
c) \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{18}\right):12\frac{1}{9}=0\)
\(\frac{1}{2}-\left(\frac{59}{9}+x-\frac{117}{18}\right):\frac{109}{9}=0\)
\(\frac{1}{2}.\left(\frac{59}{9}-\frac{117}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0:\frac{9}{109}\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0\)
\(\frac{1}{18}+x=0:\frac{1}{2}\)
\(\frac{1}{18}+x=0\)
\(x=0-\frac{1}{18}\)
\(x=\frac{-1}{18}\)
Vậy \(x=\frac{-1}{18}\)
d) 720 : [ 41 - ( 2x - 5 ) ] = 210
41 - 2x + 5 = 720 : 210
41 + 5 - 2x = \(\frac{24}{7}\)
46 - 2x = \(\frac{24}{7}\)
2x = \(46-\frac{24}{7}\)
2x = \(\frac{298}{7}\)
x = \(\frac{298}{7}:2\)
x = \(\frac{149}{7}\)
Vậy \(x=\frac{149}{7}\)

Số số hạng là :
(2x - 2) : 2 + 1 = x - 1 + 1 = x (số)
Tổng là :
(2x + 2).x : 2 = 210
=> (2x2 + 2x) : 2 = 210
=> x2 + x = 210
=> x(x + 1) = 210
=> x(x + 1) = 20.21
=> x = 20
Vậy x = 20
Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)
=> x(x + 1) = 10.2
=> x(x + 1) = 20
=> sai đề

\(15.\left(x+2\right)-12.\left(x-5\right)+21x=210\)
\(15x+30-12x+60+21x=210\)
\(24x+90=210\)
\(24x=120\)
\(x=5\)

1 + 2 + 3 + ... + x = 210
Áp dụng tính chất ta có:
1+2+3+...+x = (x+1).x:2=210
=> (x+1).x = 210 . 2
=>(x+1).x = 420
=>(x+1).x = 21 . 20
Vậy x = 20

71 + 52,5 x 4 = \(\dfrac{x+140}{x}\)+ 210
71 + 210 = (x + 140) : x + 210
71 = x : x + 140 : x (cùng bỏ trừ đi 210)
71 = 1 + 140 : x
71 - 1 = 140 : x
70 = 140 : x
140 : 70 = x
2 = x
đặt A=1+2+3+...+x
số số hạng của tổng A là: (x-1):1+1=x(số hạng)
tổng A có kết quả là: (x+1).x:2=(x+1).x/2
suy ra A=210 hay (x+1).x/2=210
suy ra x.(x+1)=210.2
x.(x+1)=420
x.(x+1)=20.21
=> x=20 (tm)
Vậy x=20
Ta có: 1+2+3+...+x=210
=>\(\frac{x\left(x+1\right)}{2}=210\)
=>\(x\left(x+1\right)=420\)
mà 420=20x21
nên x=20