\(\frac{x}{7}=\frac{x+1}{14}\)
\(\Rightarrow x\times14=\left(x+1\right)\times7\)
\(x\times14=x\times7+7\)
\(x\times14-x\times7=7\)
\(x\times\left(14-7\right)=7\)
\(x\times7=7\)'
\(\Rightarrow x=1\)
Ta có: \(\frac{x}{7}=\frac{x+1}{14}\)
=> \(\frac{2x}{14}=\frac{x+1}{14}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
x/7 = x + 1/14
x . 14 = 7x + 7
x . 14 - 7x = 7
7x = 7
x = 7 : 7
x = 1
\(\frac{x}{7}=\frac{x+1}{14}\)
\(\Rightarrow\frac{2x}{14}=\frac{x+1}{14}\)
\(=>2x=x+1\)
\(=>x=1\)
\(\frac{x}{7}=\frac{x+1}{14}\)
\(\Leftrightarrow\frac{2x}{14}=\frac{x+1}{14}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow2x-x=1\)
\(\Leftrightarrow x=1\)
Vậy x = 1
