Question

[tran thu can]

tìm x

3+6+9+12+ ... + x =693

Answer 1

[Phạm Tuấn Đạt]

\(3+6+9+...+x=693\)

\(\Rightarrow\frac{\left(x+3\right).\left[\left(x-3\right):2+1\right]}{2}=693\)

\(\Rightarrow\left(x+3\right)\left[\left(x-3\right):2+1\right]=1386\)

\(\Rightarrow\left(x+3\right)\left[\left(\frac{x}{2}-\frac{3}{2}\right)+1\right]=1386\)

\(\Rightarrow\left(x+3\right)\left[\left(\frac{x-3}{2}\right)+\frac{2}{2}\right]=1386\)

\(\Rightarrow\left(x+3\right)\left(\frac{x-1}{2}\right)=1386\)