(x+1)+(x+2)+(x+3)+(x+4)+(x+5)...+(x+15)=390

(x+x+x+x+....+x)+(1+2+3+4+5+.....+15)=390

Tổng của các số từ 1 đến 15 là:

           (15+1)x15:2=120

Có 15 số x.

Vậy Xx15+120=390

          Xx15=390-120

        Xx15=270

         X=270:15

        X=18

X x 15 + ( 1+2+3+4+....+15) =390

  vì 1+2+3+4+.....+15=120

nên 390 - 120 = 270 

x = 270 : 15 = 18 

x = 18 

mình đang âm điểm 

17 . 2 mu 4 - 15 . 2 mu 4

Các bước giải chi tiết 

a)  \(\dfrac{392-x}{32}\) + \(\dfrac{390-x}{34}\) + \(\dfrac{388-x}{36}\) = -3

⇔ \(\dfrac{392-x}{32}\)+1+\(\dfrac{390-x}{34}\)+1+\(\dfrac{388-x}{36}\)+1 = 0 

⇔\(\dfrac{424-x}{32}\)+\(\dfrac{424-x}{34}\)+\(\dfrac{424-x}{36}\)=0

⇔\(\left(424-x\right)\)\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\right)\)=0

⇔\(424-x\) = 0\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\ne\forall x\right)\)

⇔\(x=424\)

b)  \(\dfrac{x-3}{3}\)- \(x\) = \(5-\dfrac{x+1}{4}\)

⇔\(\dfrac{x-3-3x}{3}\) = 5 + \(\dfrac{-\left(x+1\right)}{4}\)

⇔\(\dfrac{-2x-3}{3}\) = 5 + \(\dfrac{-x-1}{4}\)

⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{20-x-1}{4}\)

⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{-x+19}{4}\)​​

⇔ \(4\left(-2x-3\right)\) = \(3\left(-x+19\right)\)

⇔\(-8x-12\) = \(-3x+57\)

⇔\(-8x\) = \(-3x+69\)

⇔\(-5x=69\)

⇔ \(x=-\dfrac{69}{5}\)

(392-x)/32+(390-x)/34+(388-x)/36=-3
=>392-x)/32 +1 + (390-x)/34 +1 +(388-x)/36 +1=0
=>(424-x)/32+(424-x)/34+(424-x)/36=0
=>424-x=0(vì 1/32+1/34+1/36 khác 0)
=>x=424

a, \(390-\left(x-7\right)=13^2:12\)

\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)

\(x-7=390-\dfrac{169}{12}\)

\(x-7=\dfrac{4511}{12}\)

\(x=\dfrac{4511}{12}+7\)

\(x=\dfrac{4595}{12}\)

Vậy ...

b, \(\left(x-35.2^2\right):7=3^3-24\)

\(\left(x-35.4\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(\Leftrightarrow\left(x-140\right)=3.7\)

\(\Leftrightarrow x-140=21\)

\(\Leftrightarrow x=161\)

Vậy .....

c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)

\(x-3-\left(16.3-24\right):2:6=3\)

\(x-3-\left(48-24\right):2:6=3\)

\(x-3-24:2:6=3\)

\(x-3-2=3\)

\(x=3+2+3\)

\(x=8\)

Vậy ......

d) \(4x-5=5+5^2+5^3+.....+5^{99}\)

Đặt :

\(A=5+5^2+.........+5^{99}\)

\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)

\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)

\(\Leftrightarrow4A=5^{100}-5\)

\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)

\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)

Đến đây thì sao nữa nhỉ ?

e) \(\left(2x-1\right)^4=625\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy ....

a) x(x+1)=2+4+6+8+...+2500

x(x+1)=(2500+2).1250:2

x(x+1)=1563750

Ta thấy: x và x+1 là 2 số liên tiếp

Mà x(x+1)=1563750=>x=1250;x+1=1251

Vậy x=1250

b) x-3=12:{390:[500-(5^3+7^2.5)]}

x-3=12:[390:(500-254.64)]

x-3=12:(390:245.35)

x-3=12:1.58

x-3=7.54

=>x=10.5

Vậy x=10.5

c) x+3^3.12=3^3.19

=>x=2.46

Vậy x=2.46

(x+1) + (x+2) + (x+3) +...+ (x+20) = 390

x + 1 + x + 2 + x + 3 + ... + x + 20 = 390

20x + ( 1 + 2 + 3 + .. +20 ) = 390

                         20x + 210 = 390

                         20x          = 390 - 210

                         20x          = 180

                            x           = 180 : 20

                            x           = 9

Vậy x = 9

( x -1 ) + ( x -2 ) + ( x - 3 ) + ... + ( x - 20 ) = 390

x - 1 + x -2 + x - 3 + ... + x - 20 = 390

20x - ( 1 + 2 + 3 + ... + 20 ) = 390

20x - 210 = 390

20x         = 390 + 210 

20x         = 600

x             = 600 : 20

x             = 30

Vậy x= 30

P/S:Bài đầu mik làm sai đề ^_^

suy ra (x+x+x+...+x)-(1+2+3+...+20)=390 (1)

do (1+2+3+...+20) có 20 số hạng

suy ra x+x+x+...+x có 20 số hạng

suy ra x+x+x+...+x=20x (2)

ta có 1+2+3+...+20=(1+20)x20:2=210 (3)

từ (1),(2),(3) suy ra 20x+210=390

                               20x=180

                                   x=6