\(\frac{x+1}{x+2}=\frac{0,8}{1,2}\Leftrightarrow\frac{x+1}{x+2}=\frac{2}{3}\)

Áp dụng t/c tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)

Ta có: \(\left(x+1\right).3=\left(x+2\right).2\)

\(\Rightarrow3x+3=2x+4\Rightarrow3x-2x=4-3\Rightarrow x=1\)

Vậy x=1