Question

Tìm x, biết x là số tự nhiên và:

 (x+1)+(x+2)+(x+3)+........+(x+100)=5050

Answer 1

\(\left[x+1\right]+\left[x+2\right]+\left[x+3\right]+...+\left[x+100\right]=5050\)

\(\Rightarrow x+1+x+2+x+3+...+x+100=5050\)

\(\Rightarrow(x+x+...+x)+(1+2+3+...+100)=5050\)

\(\Rightarrow100x+5050=5050\)

\(\Rightarrow100x=5050-5050\)

\(\Rightarrow100x=0\)

\(\Rightarrow x=\frac{0}{100}=0\)

Answer 2

\(\Leftrightarrow100x+5050=5050\Rightarrow x=0\)

Answer 3

(x+1)+(x+2)+(x+3)+...........+(x+100)=5050

x+1+x+2+x+3+.............+x+100=5050

x*100+1+2+3+.............+100=5050

x*100+5050=5050

x*100=5050-5050=0

x=0/100=0

Answer 4

 (x+1)+(x+2)+(x+3)+........+(x+100)=5050

=> x+1+x+2+x+3+.....+x+100=5050

=>(x+x+x+.....+x)+(1+2+3+....+100)=5050

      (100 c/s x)         (100 số hạng )

=>100x +  5050=5050

=> 100x          = 5050-5050

=>100x       = 0

=> x= 0 : 100

=> x = 0

Vậy x=0

T I C K giùm mik

Answer 5

trả lời 

x=0/100=>x=0

chúc bn 

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