\(\left(x+5\right)+\left(x+10\right)+\left(x+15\right)+...+\left(x+100\right)=1110\)
\(\Leftrightarrow x+5+x+10+x+15+...+x+100=1110\)
\(\Leftrightarrow20x+\left(5+10+15+...+100\right)=1110\)
\(\Leftrightarrow20x+\frac{\left(100+5\right)\left[\left(100-5\right)\div5+1\right]}{2}=1110\)
\(\Leftrightarrow20x+\frac{2100}{2}=1110\)
\(\Leftrightarrow20x+1050=1110\)
\(\Leftrightarrow20x=1110-1050\)
\(\Leftrightarrow20x=60\)
\(\Leftrightarrow x=60\div20\)
\(\Leftrightarrow x=3\)
Vậy x = 3
\((x+5)+(x+10)+(x+15)+...+(x+100)=1110\)
\(\Rightarrow x+5+x+10+x+15+...+x+100=1110\)
\(\Rightarrow(x+x+x+...+x)+(5+10+15+...+100)=1110\)
\(\Rightarrow100x+\frac{(5+100)\cdot20}{2}=1110\)
\(\Rightarrow100x+1050=1110\)
\(\Rightarrow100x=1110-1050\)
\(\Rightarrow100x=60\)
....
Ta có (x+5)+(x+10)+(x+15)+....+(x+100)=1110
=>(x×100)+(5+10+15+...+100)=1110
=>(x×100)+ 1050=1110
=> x×100= 1110-1050=60
=>x= 3/5
Vậy ...
(x+5)+(x+10)+(x+15)+...+(x+100)=1110
=>x+5+x+10+x+15+...+x+100=1110
=>(x+x+x+...+x)+(5+10+15+...+100)=1110
=>20x+[(100-5):5+1].(100+5):2=1110
=>20x+20.105:2=1110
=>20x+2100:2=1110
=>20x+1050=1110
=>20x=1110-1050=60
=>x=60:20=3
Vậy x=3
~hok tốt~
\(\left(x+5\right)+\left(x+10\right)+\left(x+15\right)+....+\left(x+100\right)=1110\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(5+10+15+....+100\right)=1110\)
\(\Rightarrow20x+1050=1110\)
\(\Rightarrow\) \(20x=1110-1050\)
\(\Rightarrow\) \(20x=60\)
\(\Rightarrow\) \(x=60\div20\)
\(\Rightarrow\) \(x=3\)
\((x+5)+(x+10)+(x+15)+...+(x+100)=1110\)
\(\Rightarrow x+5+x+10+x+15+...+x+100=1110\)
\(\Rightarrow(x+x+x+...+x)+(5+10+15+...+100)=1110\)
\(\Rightarrow20x+1050=1110\)
\(\Rightarrow20x=1110-1050\)
\(\Rightarrow20x=60\)
\(\Rightarrow x=3\)
Làm lại , tự nhận mình ra cái sai :<<
