\(\Rightarrow\dfrac{\left(x+100+x\right)\left(x+100-x+1\right)}{2}=10100\\ \Rightarrow101\left(2x+100\right)=20200\\ \Rightarrow2x+100=200\Rightarrow x=50\)

<=>     101x+(1+99)+(2+98)+....+(49+51)+100+50=10100
<=>      101x+5050=10100
<=>      101x=5050
<=>            x=50

\(x+\left(x+1\right)+...+\left(x+100\right)=10100\)

\(\Leftrightarrow101x+5050=10100\)

\(\Leftrightarrow101x=5050\)

hay x=50

x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 10100

( x + x + x + ... + x ) + ( 1 + 2 + .. + 100 ) =10100

101x       + 5050                                     =10100

101x                                                      =5050

x                                                            =50

x + (x+1) + (x+2) + ... + (x+100) = 10100

=> 101x + (1 + 2 + 3 + ... + 100) = 10100

Áp dụng công thức tính dãy số ta có :

1 + 2 + 3 +...+ 100 = \(\frac{\left[\left(100-1\right):1+1\right].\left(100+1\right)}{2}=\frac{100.101}{2}=50.101=5050\)

=> 101x + 5050 = 10100

=> 101x = 5050

=> x = 50

x = 50 nha

x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 10100

x + x + 1 + x + 2 + ... + x + 100 = 10100

( x + x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 10100

101x + 5050 = 10100

101x = 5050

x = 50

Vậy x = 50

X + (X + 1) + (X + 2) + .... + (X + 100) = 10100

X + (X + X + ... + X) + (1 + 2 + .... + 100) = 10100

X + X x 100 + 5050 = 10100

X x 101 = 10100 - 5050

X x 101 = 5050

X           = 5050 : 101

X           = 50

x+(x+1)+(x+2)+...+(x+100)-10100(thiếu dấu ngoặc)

\(M=\frac{9^2}{9.10}.\frac{10^2}{10.11}.\frac{11^2}{11.12}.....\frac{100^2}{100.101}\)

\(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{100}{101}=\frac{9}{101}\)