\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=5950\)

\(x+x+1+x+2+...+x+99=5950\)

\(100x+\left(1+2+3+...+99\right)=5950\)

\(100x+\left(99+1\right).99:2=5950\)

\(100x+4950=5950\)

\(100x=5950-4950\)

100x=1000

x=10

x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ..+ ( x + 99 ) = 5950

=> ( x + x + x +....+x) + ( 1 + 2 + 3+ ....+ 99) = 5950

Có tất cả số hạng ở dãy số tự nhiên là

( 99 -1) : 1 + 1 = 99 số

Tổng dãy số tự nhiên là

( 99 + 1) x 99 : 2 = 4950

Tổng dãy số x là

99 + 1=100 số x

 x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ..+ ( x + 99 ) = 5950

hay 100x + 4950 = 5950

100x = 5950 - 4950

100x = 1000

x = 10

Vậy x =10

( x + 100 )  + ( x + 99 ) + ( x + 98 ) + ........ + ( x + 2 ) + ( x + 1 ) = 5950

=> (x+x+x+.......+x)+(1+2+3+............+99+100)=5950

=> 100x+(100+1).100:2=5950

=> 100x+10100:2=5950

=>100x+5050=5950

=> 100x=5950-5050

=> 100x=900

=> x=900:100

=> x=9

( x + 100 )  + ( x + 99 ) + ( x + 98 ) + ........ + ( x + 2 ) + ( x + 1 ) = 5950

( x + x + x + ... + x + x ) + ( 100 + 99 + 98 + ... + 2 + 1 ) = 5950

100x + 5050 = 5950

100x = 5950 - 5050

100x = 900

x = 900 : 100

x = 9

1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)

\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)

\(\Rightarrow3S-S=3^{2021}-1\)

\(\Rightarrow2S=3^{2021}-1\)

\(\Rightarrow S=\frac{3^{2021}-1}{2}\)

2. \(\left(3x-2\right)^3=64\)

\(\Leftrightarrow\left(3x-2\right)^3=4^3\)

\(\Leftrightarrow3x-2=4\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

[3x-2]^3=64

Ta có:64=4^3    

Suy ra:3x-2=4

           3x   =4+2

          3x=6

         x=6:3

         x=2

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

(100 - 99 + 98 - 87 + ... + 2 - 1) : 50 + 2010 - 12x = 21

=> (1 + 1 + ... + 1) : 50 + 2010 - 12x = 21

=> 50 : 50 + 2010 - 12x = 21

=> 1 + 2010 - 12x = 21

=> 2011 - 12x = 21

=> 12x = 2011 - 21

=> 12x = 1990

=> x = 995/6

dúng không bạn 

ta có 100 - 99 + 98 -97+ ...+ 4 - 3 + 2 - 1 = (100 - 99) + (98 - 97) + ...+ (4 - 3) + (2 - 1)

= 1+1+1+..+1+1( 50 chữ số 1)

= 1 x 50 = 50

theo đề ra ta có 50 : 50 + 2010 - 12 * x = 21

=> 2013 - 12*x = 21

=> 12*x = 2013 - 21 = 1992

=> x = 1992 : 12 = 166

Câu b,

= 50x + (1 + 3 + 5 + ..... + 99) = 2550

= 50x + 2500 = 2550

= 50x = 2550 - 2500

= 50x = 50

= x = 50 : 50

= x = 1

refer

A) 2448:[119-[x-6]]=24

<=> 199 - ( x-6 ) = 2448 : 24 = 102

<=> x - 6 = 199 - 102 = 97

<=> x = 97 +6 = 103

B) (x+1)+(x+3)+(x+5)+......+(x+99) = 2550

(x+x+x+....+x)+(1+3+5+....+99) = 2550

50x + 2500 = 2550

50x = 2550 - 2500

50x = 50

x = 50 : 50

x = 1

viết tập hợp à bạn?

tìm x eN nhé các bạn

\(A=1.2.3+2.3.4+...+98.99.100\)

\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+....+98.99.100.\left(101-97\right)\)

\(4A=1.2.3.4+2.3.4.5-1.2.3.4+...+98.99.100.101-97.98.99.100\)

\(4A=98.99.100.101\)

\(A=\frac{98.99.100.101}{4}\)

\(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|\)

=> \(\left|x+1\right|\ge0\)tt 

\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\\x+5=0\end{cases}}\Rightarrow x=\hept{\begin{cases}x=-1\\x=-3\\x=-5\end{cases}}\)

thoả mãn giá trị trên