Question

\(|\frac{3-2x}{1+x}|>4\)

giải bpt

Answer 1

ĐKXĐ : \(x\ne-1\)

\(\left|\frac{3-2x}{1+x}\right|>4\)\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{3-2x}{1+x}>4\left(1\right)\\\frac{2x-3}{1+x}< -4\left(2\right)\end{cases}}\)

\(\left(1\right)\)\(\Leftrightarrow\)\(3-2x>4+4x\)\(\Leftrightarrow\)\(x< \frac{-1}{6}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(2x-3< -4-4x\)\(\Leftrightarrow\)\(x< \frac{-1}{6}\)

Vậy \(x< \frac{-1}{6}\)

PS : ko wen làm pt nên sai sót thì bỏ qua nhé :)