(x - 1) + (x - 2) + (x - 3) + ... + (x - 100) = 50

(x + x + x + ... +x) - (1 + 2 + 3 + ... + 100) = 50

100x - 5050 = 50

100x = 50 + 5050

100x = 5100

x = 5100 : 100 = 51

(x + 1) + (x + 2 ) + (x + 3) + ... + (x - 100) = 50 
x - 1 + x - 2 + x - 3 + .... + x - 100 = 50 
(x + x + x + ... + x) + (1 - 2 - 3 - ... - 100) = 50 
100x + (-2475) = 5750 
100x = 5750 - (-2475) =  8225
x = 8225 : 100 = 82,25

Số số hạng từ 1 đến 100 là :(100-1):1+1=100(số hạng)

tổng :(100+1).100:2=5050

=>có 100 x.Ta có:5050+100.x=50

                                       100.x=5050-50

                                        100.x=5000

                                               x=5000:100

                                              x=50

                                     vậy x=50

x+100=50x3

x+100=150

x=150-100

x=50

tk mk mk tk lại

x + 100 = 50 x 3

x + 100 =   150

x           = 150 - 100

x           =  50

Tìm x :

x + 100 = 50 x 3

x + 100 = 150

x          = 150 - 100

x          = 50

Vậy x= 50

` (3/4 +x) xx 50/100 = 8/10`

`( 3/4 + x) xx 1/2 = 4/5`

` 3/4 + x = 4/5 : 1/2`

` 3/4 + x = 4/5 xx2`

` 3/4 + x = 8/5`

` x = 8/5 - 3/4`

` x = 17/20`

Tìm x, biết:

( 3/4 + x ) x 50/100 = 8/10

( 3/4 + x ) = 8/10 : 50/100

( 3/4 + x ) = 8/5

x = 8/5 - 3/4

x = 17/20.

=>x+3/4=4/5:1/2=4/5x2=8/5

=>x=8/5-3/4=32/20-15/20=17/20

=> X x ( 1+2+3+4+....+50) = 2 x ( 1+2+3+...+50)

=> x=2

(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=50

100x+5050=50

100x=-5000

x=-50

a; 100% : \(x\) - 50% : \(x\) + 40% : \(x\) = 18 + 30% : \(x\)

   (100% - 50% + 40%): \(x\) = 18 + 30% : \(x\)

                            90% : \(x\)  = 18 + 30% : \(x\)                  

                           90%: \(x\)  - 30%: \(x\) = 18

                          60% : \(x\) = 18

                                   \(x\) = 60% : 18

                                    \(x\) = \(\dfrac{1}{30}\)

x²+x³+x⁴+...+x¹⁰⁰=50

Có số số hạng lũy thừa x là: 100-2+1=99(số)

Có số nhóm 3 ghép thành là:99:3=33

Ta có:x²+x³+x⁴+...+x¹⁰⁰=50

<=>(x²+x³+x⁴)+...+(x⁹⁸+x⁹⁹+x¹⁰⁰)=50

ôi suy nghĩ cái

\(x+3x+3^2x+3^3x+...+3^{49}x=3^{100}-3^{50}\)

\(\Leftrightarrow x\left(1+3+3^2+3^3+...+3^{49}\right)=3^{100}-3^{50}\)

Đặt \(A=1+3+3^2+3^3+...+3^{49}\)

\(\Leftrightarrow3A=3+3^2+3^3+3^4+3^{50}\)

\(\Leftrightarrow3A-A=2A=3^{50}-1\)

\(\Leftrightarrow A=\frac{3^{50}-1}{2}\)

\(\Rightarrow\frac{x\left(3^{50}-1\right)}{2}=3^{100}-3^{50}\)

\(\Leftrightarrow x\left(3^{50}-1\right)=2.3^{100}-2.3^{50}\)

\(\Leftrightarrow x=\frac{2.3^{100}-2.3^{50}}{3^{50}-1}\)

\(x+3x+3^2x+3^3x+...+3^{49}x=3^{100}-3^{50}\)

\(\Leftrightarrow x\left(1+3+3^2+...+3^{49}\right)=3^{100}-3^{50}\)

\(\Leftrightarrow x\left(\frac{3^{50}-1}{2}\right)=3^{100}-3^{50}\)

\(\Leftrightarrow x\left(3^{50}-1\right)=3^{102}-3^{52}\)

\(\Rightarrow x=\frac{3^{102}-3^{52}}{3^{50}-1}=\frac{3^{52}\left(3^{50}-1\right)}{3^{50}-1}=3^{52}\)