1188      

Đáp án cần chọn là: B

x = 0.

100x+5050=102x

100x+5050=100x+2x

=>2x=5050

x=5050/2

x=2525

(x + 1) + (x + 2) + ... + (x + 100) = 102x

x + x + x + ... + x + 1 + 2 + ... + 100 = 102x

100x + (100 + 1) x 100 : 2 = 102x

102x - 100x = 5050

2x = 5050

x = 2525

( x + 1 ) + ( x+ 2 ) + ... + ( x + 99 ) = 0

<=> 99x + ( 1+2+3+..+99) = 0

<=> 99x + ( 100 . 99 : 2 ) =0

<=> 99x + 4950=0

<=> 9x = -4950

<=> x = -4950 : 9

<=> x = -550

mình hỏi xem cách lm mình có đúng ko

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

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

x.100 + 4950 = 99

x.100 = 99 - 4950

x = - 4851 : 100

x = - 48,51

ta có x +(x+1)+(x+2)+......+(x+99)=99

        100x + (1+2+3+...+99)=99

        100x + 4950 = 99

        100x = -4851

              = -48,51

Xét tổng 1 + 3 +5 +... + 99:

- Số số hạng là: (99-1):2+1= 50 số

- Tổng là: (1 + 99) x 50 : 2 = 2500

Vậy ta có biểu thức:

 \(2500=\left(x-2\right)^2\\ \Leftrightarrow50^2=\left(x-2\right)^2\\\Leftrightarrow \left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

Vậy...

\(1+3+5+...+99=\left(x-2\right)^2\)

\(\Rightarrow\left[\left(99-1\right):2+1\right]\left(1+99\right):2=\left(x-2\right)^2\)

\(\Rightarrow50.100:2=\left(x-2\right)^2\)

\(\Rightarrow\left(x-2\right)^2=2500=50^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)