Question

 \(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+...+\left(x+100\right)=11000\)

Answer 1

x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ... + ( x + 100 ) = 11 000

<=> ( x + x + x + x + x + ... + x ) + ( 1 + 2 + 3 + 4 + ... + 100 ) = 11 000

<=> 101x +  \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 11 000 ( vì sao em để 101x thì idol biết mà :33 )

<=> 101x + 5050 = 11 000

<=> 101x = 5950

<=> x = 5950/101

Answer 2

x + (x + 1) + (x + 2) + ......+ (x + 100) = 11000

x +( x + x + ...... + x ) + (1 + 2 + ...... + 100) = 11000

x + 100x + 5050 = 11000

x + 100x = 5950

101x = 5950

x = 5950 : 101

x = 5950 : 100 + 5950 : 1

x = 59,50 + 5950

x = 6009,50

Answer 3

\(x+\left(x+1\right)+\left(x+2\right)+.................+\left(x+100\right)=11000\)

\(x+x+1+x+2+x+3+..............+x+100=11000\)

\(\left(x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=11000\)

Có 101 số hạng x                                         Có 100 số hạng

\(101x+\left[\left(1+100\right).100:2\right]=11000\)

\(101x+\left(101.100:2\right)=11000\)

\(101x+5050=11000\)

\(101x=11000-5050\)

\(101x=5950\)

\(x=\frac{5950}{101}\)

Vậy \(x=\frac{5950}{101}\)

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